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CHAPTER

1

The Nature of Physical Chemistry and the Kinetic Theory of Gases

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition Publisher: MCH Multimedia inc.

Problems and Solutions

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Classical Mechanics and Thermal Equilibrium

Chapter 1 *problems with an asterisk are slightly more demanding Classical Mechanics and Thermal Equilibrium 1.1.

Calculate the amount of work required to accelerate a 1000-kg car (typical of a Honda Civic) to 88 km hr–1 (55 miles hr–1). Compare this value to the amount of work required for a 1600-kg car (typical of a Ford Taurus) under the same conditions. Solution

1.2.

Assume that a rod of copper is used to determine the temperature of some system. The rod’s length at 0 °C is 27.5 cm, and at the temperature of the system it is 28.1 cm. What is the temperature of the system? The linear expansion of copper is given by an equation of the form lt = l0(1 + αt + βt2) where α = 0.160 × 10–4 K–1, β = 0.10 × 10–7 K–2, l0 is the length at 0 °C, and lt is the length at t °C. Solution

1.3.

Atoms can transfer kinetic energy in a collision. If an atom has a mass of 1 × 10–24 g and travels with a velocity of 500 m s–1, what is the maximum kinetic energy that can be transferred from the moving atom in a head-on elastic collision to the stationary atom of mass 1 × 10–23 g? Solution

1.4.

Power is defined as the rate at which work is done. The unit of power is the watt (W = 1 J s–1). What is the power that a man can expend if all his food consumption of 8000 kJ a day (≈ 2000 kcal) is his only source of energy and it is used entirely for work? Solution

1.5.

State whether the following properties are intensive or extensive: (a) mass; (b) density; (c) temperature; (d) gravitational field. Solution

1-2

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Gas Laws and Temperature

Gas Laws and Temperature 1.6.

The mercury level in the left arm of the J-shaped tube in Fig. 1.6a is attached to a thermostat gas-containing bulb. The left arm is 10.83 cm and the right arm is 34.71 cm above the bottom of the manometer. If the barometric pressure reads 738.4 Torr, what is the pressure of the gas? Assume that temperature-induced changes in the reading of the barometer and J tube are small enough to neglect. Solution

1.7.

Vacuum technology has become increasingly more important in many scientific and industrial applications. The unit Torr, defined as 1/760 atm, is commonly used in the measurement of low pressures. a. Find the relation between the older unit mmHg and the Torr. The density of mercury is 13.5951 g cm–3 at 0.0 °C. The standard acceleration of gravity is defined as 9.806 65 m s–2. b. Calculate at 298.15 K the number of molecules present in 1.00 m3 at 1.00 × 10–6 Torr and at 1.00 × 10–15 Torr (approximately the best vacuum obtainable). Solution

1.8.

The standard atmosphere of pressure is the force per unit area exerted by a 760-mm column of mercury, the density of which is 13.595 11 g cm–3 at 0 °C. If the gravitational acceleration is 9.806 65 m s–2, calculate the pressure of 1 atm in kPa. Solution

1.9.

Dibutyl phthalate is often used as a manometer fluid. Its density is 1.047 g cm–3. What is the relationship between 1.000 mm in height of this fluid and the pressure in torr? Solution

1.10.

The volume of a vacuum manifold used to transfer gases is calibrated using Boyle’s law. A 0.251-dm3 flask at a pressure of 697 Torr is attached, and after system pumpdown, the manifold is at 10.4 Torr. The stopcock between the manifold and flask is opened and the system reaches an equilibrium pressure of 287 Torr. Assuming isothermal conditions, what is the volume of the manifold? Solution

1.11. An ideal gas occupies a volume of 0.300 dm3 at a pressure of 1.80 × 105 Pa. What is the new volume of the gas maintained at the same temperature if the pressure is reduced to 1.15 × 105 Pa? Solution 1-3

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.12.

Gas Laws and Temperature

If the gas in Problem 1.11 were initially at 330 K, what will be the final volume if the temperature were raised to 550 K at constant pressure? Solution

1.13.

Calculate the concentration in mol dm–3 of an ideal gas at 298.15 K and at (a) 101.325 kPa (1 atm), and (b) 1.00 × 10–4 Pa (= 10–9 atm). In each case, determine the number of molecules in 1.00 dm3. Solution

*1.14. A J-shaped tube is filled with air at 760 Torr and 22 °C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a funnel into the open end. When the mercury spills over the top of the short arm, what is the pressure on the trapped air? Let h be the length of mercury in the long arm. Solution 1.15.

A Dumas experiment to determine molar mass is conducted in which a gas sample’s P, θ, and V are determined. If a 1.08-g sample is held in 0.250 dm3 at 303 K and 101.3 kPa: a. What would the sample’s volume be at 273.15 K, at constant pressure? b. What is the molar mass of the sample? Solution

1.16.

A gas that behaves ideally has a density of 1.92 g dm–3 at 150 kPa and 298 K. What is the molar mass of the sample? Solution

1.17. The density of air at 101.325 kPa and 298.15 K is 1.159 g dm–3. Assuming that air behaves as an ideal gas, calculate its molar mass. Solution 1.18.

A 0.200-dm3 sample of H2 is collected over water at a temperature of 298.15 K and at a pressure of 99.99 kPa. What is the pressure of hydrogen in the dry state at 298.15 K? The vapor pressure of water at 298.15 K is 3.17 kPa. Solution

1-4

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.19.

Gas Laws and Temperature

What are the mole fractions and partial pressures of each gas in a 2.50-L container into which 100.00 g of nitrogen and 100.00 g of carbon dioxide are added at 25 °C? What is the total pressure? Solution

1.20.

The decomposition of KClO3 produces 27.8 cm3 of O2 collected over water at 27.5 °C. The vapor pressure of water at this temperature is 27.5 Torr. If the barometer reads 751.4 Torr, find the volume the dry gas would occupy at 25.0 °C and 1.00 bar. Solution

1.21.

Balloons now are used to move huge trees from their cutting place on mountain slopes to conventional transportation. Calculate the volume of a balloon needed if it is desired to have a lifting force of 1000 kg when the temperature is 290 K at 0.940 atm. The balloon is to be filled with helium. Assume that air is 80 mol % N2 and 20 mol % O2. Ignore the mass of the superstructure and propulsion engines of the balloon. Solution

*1.22. A gas mixture containing 5 mol % butane and 95 mol % argon (such as is used in Geiger-Müller counter tubes) is to be prepared by allowing gaseous butane to fill an evacuated cylinder at 1 atm pressure. The 40.0-dm3 cylinder is then weighed. Calculate the mass of argon that gives the desired composition if the temperature is maintained at 25.0 °C. Calculate the total pressure of the final mixture. The molar mass of argon is 39.9 g mol–1. Solution 1.23.

The gravitational constant g decreases by 0.010 m s–2 km–1 of altitude. a. Modify the barometric equation to take this variation into account. Assume that the temperature remains constant. b. Calculate the pressure of nitrogen at an altitude of 100 km assuming that sea-level pressure is exactly 1 atm and that the temperature of 298.15 K is constant. Solution

1.24.

Suppose that on another planet where the atmosphere is ammonia that the pressure on the surface, at h = 0, is 400 Torr at 250 K. Calculate the pressure of ammonia at a height of 8000 metres. The planet has the same g value as the earth. Solution

1-5

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Graham’s Law, Molecular Collisions, and Kinetic Theory

1.25. Pilots are well aware that in the lower part of the atmosphere the temperature decreases linearly with altitude. This dependency may be written as T = T0 – az, where a is a proportionality constant, z is the altitude, and T0 and T are the temperatures at ground level and at altitude z , respectively. Derive an expression for the barometric equation that takes this into account. Work to a form involving ln P/P0. Solution 1.26.

An ideal gas thermometer and a mercury thermometer are calibrated at 0 °C and at 100 °C. The thermal expansion coefficient for mercury is



1 ( V / T ) P V0

 1.817  10 4  5.90  10 9   3.45  1010  2 where θ is the value of the Celsius temperature and V0 = V at θ = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 °C? Solution Graham’s Law, Molecular Collisions, and Kinetic Theory 1.27. It takes gas A 2.3 times as long to effuse through an orifice as the same amount of nitrogen. What is the molar mass of gas A? Solution 1.28.

Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, takes 5.80 minutes to effuse through an orifice. How long will it take for helium to effuse under the same conditions? Solution

1.29. What is the total kinetic energy of 0.50 mol of an ideal monatomic gas confined to 8.0 dm3 at 200 kPa? Solution 1.30.

Nitrogen gas is maintained at 152 kPa in a 2.00-dm3 vessel at 298.15 K. If its molar mass is 28.0134 g mol–1 calculate: a. The amount of N2 present. b. The number of molecules present. 1-6

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Graham’s Law, Molecular Collisions, and Kinetic Theory

c. The root-mean-square speed of the molecules. d. The average translational kinetic energy of each molecule. e. The total translational kinetic energy in the system. Solution 1.31. By what factor are the root-mean-square speeds changed if a gas is heated from 300 K to 400 K? Solution *1.32. The collision diameter of N2 is 3.74 × 10–10 m at 298.15 K and 101.325 kPa. Its average speed is 474.6 m s–1. Calculate the mean free path, the average number of collisions ZA experienced by one molecule in unit time, and the average number of collisions ZAA per unit volume per unit time for N2. Solution *1.33. Express the mean free path of a gas in terms of the variables pressure and temperature, which are more easily measured than the volume. Solution 1.34.

Calculate ZA and ZAA for argon at 25 °C and a pressure of 1.00 bar using d = 3.84 × 10–10m obtained from X-ray crystallographic measurements. Solution

1.35.

Calculate the mean free path of Ar at 20 °C and 1.00 bar. The collision diameter d = 3.84 × 10–10 m. Solution

1.36.

Hydrogen gas has a molecular collision diameter of 0.258 nm. Calculate the mean free path of hydrogen at 298.15 K and (a) 133.32 Pa, (b) 101.325 k Pa, and (c) 1.0 × 10 8 Pa. Solution

1.37. In interstellar space it is estimated that atomic hydrogen exists at a concentration of one particle per cubic meter. If the collision diameter is 2.5 × 10–10 m, calculate the mean free path λ. The temperature of interstellar space is 2.7 K. Solution 1-7

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Distributions of Speeds and Energies

*1.38. Calculate the value of Avogadro’s constant from a study made by Perrin [Ann. Chem. Phys., 18, 1(1909)] in which he measured as a function of height the distribution of bright yellow colloidal gamboge (a gum resin) particles suspended in water. Some data at 15 °C are: height, z/10–6 N, relative number of gamboge particles at height z

5

35

100

47

–3

ρgamboge = 1.206 g cm

ρwater = 0.999 g cm-3 radius of gamboge particles, r = 0.212 × 10–6 m (Hint: Consider the particles to be gas molecules in a column of air and that the number of particles is proportional to the pressure.) Solution Distributions of Speeds and Energies 1.39.

Refer to Table 1.3 (p. 32) and write expressions and values for (a) the ratio u 2 / u , and (b) the ratio ū/ump. Note that these ratios are independent of the mass and the temperature. How do the differences between them depend on these quantities? Solution

1.40.

The speed that a body of any mass must have to escape from the earth is 1.07 × 104 m s–1. At what temperature would the average speed of (a) an H2 molecule, and (b) an O2 molecule be equal to this escape speed? Solution

1.41.

a. For H2 gas at 25 °C, calculate the ratio of the fraction of molecules that have a speed 2u to the fraction that have the average speed ū. How does this ratio depend on the mass of the molecules and the temperature? b. Calculate the ratio of the fraction of the molecules that have the average speed ū100ºC at 100 °C to the fraction that have the average speed ū25ºC at 25 °C. How does this ratio depend on the mass? Solution

1-8

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Real Gases

1.42. Suppose that two ideal gases are heated to different temperatures such that their pressures and vapor densities are the same. What is the relationship between their average molecular speeds? Solution 1.43.

a. If ū25 ºC is the average speed of the molecules in a gas at 25 °C, calculate the ratio of the fraction that will have the speed ū25 ºC at 100° to the fraction that will have the same speed at 25 °C. b. Repeat this calculation for a speed of 10 ū25ºC. Solution

1.44.

On the basis of Eq. 1.80 with β = 1/kBT, derive an expression for the fraction of molecules in a one-dimensional gas having speeds between ux and ux + dux. What is the most probable speed? Solution

*1.45. Derive an expression for the fraction of molecules in a one-dimensional gas having energies between x and x  d x .Also, obtain an expression for the average energy x . Solution *1.46. Derive an expression for the fraction of molecules in a two-dimensional gas having speeds between u and u + du. (Hint: Proceed by analogy with the derivation of Eq. 1.91.) Then obtain the expression for the fraction having energies between and  d . What fraction will have energies in excess of *? Solution Real Gases 1.47.

In Section 1.13 it was stated that the van der Waals constant b is approximately four times the volume occupied by the molecules themselves. Justify this relationship for a gas composed of spherical molecules. Solution

1.48.

Draw the van der Waals PV isotherm over the same range of P and V as in Figure 1.21 at 350 K and 450 K for Cl2 using the values in Table 1.4. Solution 1-9

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Real Gases

1.49. Compare the pressures predicted for 0.8 dm3 of Cl2 weighing 17.5 g at 273.15 K using (a) the ideal gas equation and (b) the van der Waals equation. Solution 1.50.

A particular mass of N2 occupies a volume of 1.00 L at –50 °C and 800 bar. Determine the volume occupied by the same mass of N2 at 100 °C and 200 bar using the compressibility factor for N2. At –50 °C and 800 bar it is 1.95; at 100 °C and 200 bar it is 1.10. Compare this value to that obtained from the ideal gas law. Solution

1.51. A gas is found to obey the equation of state

P

RT a  V b V

where a and b are constants not equal to zero. Determine whether this gas has a critical point; if it does, express the critical constants in terms of a and b. If it does not, explain how you determined this and the implications for the statement of the problem. Solution 1.52.

Ethylene (C2H4) has a critical pressure of Pc = 61.659 atm and a critical temperature of Tc = 308.6 K. Calculate the molar volume of the gas at T = 97.2 °C and 90.0 atm using Figure 1.22. Compare the value so found with that calculated from the ideal gas equation. Solution

1.53.

Assuming that methane is a perfectly spherical molecule, find the radius of one methane molecule using the value of b listed in Table 1.5. Solution

1.54. Determine the Boyle temperature in terms of constants for the equation of state: PVm = RT{1 + 8/57(P/Pc)(Tc/T)[1 – 4(Tc/T)2]} R, Pc, and Tc are constants. Solution

1-10

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Real Gases

1.55. Establish the relationships between van der Waals parameters a and b and the virial coefficients B and C of Eq. 1.117 by performing the following steps: a. Starting with Eq. 1.101, show that

PVm V a 1  m  . RT Vm  b RT Vm b. Since Vm/(Vm – b) = (1 – b/Vm)–1, and (1 – x)–1 = 1 + x + x2 + …,, expand (1 – b/ Vm)–1 to the quadratic term and substitute into the result of part (a). c. Group terms containing the same power of Vm and compare to Eq. 1.117 for the case n = 1. d. What is the expression for the Boyle temperature in terms of van der Waals parameters? Solution *1.56. Determine the Boyle temperature of a van der Waals gas in terms of the constants a, b, and R. Solution 1.57.

The critical temperature Tc of nitrous oxide (N2O) is 36.5 °C, and its critical pressure Pc is 71.7 atm. Suppose that 1 mol of N2O is compressed to 54.0 atm at 356 K. Calculate the reduced temperature and pressure, and use Figure 1.22, interpolating as necessary, to estimate the volume occupied by 1 mol of the gas at 54.0 atm and 356 K. Solution

1.58.

At what temperature and pressure will H2 be in a corresponding state with CH4 at 500.0 K and 2.00 bar pressure? Given Tc = 33.2 K for H2, 190.6 K for CH4; Pc = 13.0 bar for H2, 46.0 bar for CH4. Solution

*1.59. For the Dieterici equation, derive the relationship of a and b to the critical volume and temperature. [Hint: Remember that at the critical point (∂P/∂V)T = 0 and (∂2P/∂V2)T = 0.] Solution

1-11

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.60.

Real Gases

In Eq. 1.103 a cubic equation has to be solved in order to find the volume of a van der Waals gas. However, reasonably accurate estimates of volumes can be made by deriving an expression for the compression factor Z in terms of P from the result of the previous problem. One simply substitutes for the terms Vm on the right-hand side in terms of the ideal gas law expression Vm = RT/P. Derive this expression and use it to find the volume of CCl2F2 at 30.0 °C and 5.00 bar pressure. What will be the molar volume computed using the ideal gas law under the same conditions? Solution

*1.61. A general requirement of all equations of state for gases is that they reduce to the ideal gas equation (Eq. 1.28) in the limit of low pressures. Show that this is true for the van der Waals equation. Solution 1.62. The van der Waals constants for C2H6 in the older literature are found to be a = 5.49 atm L2 mol–2 and b = 0.0638 L mol–1 Express these constants in SI units (L = liter = dm3). Solution *1.63. Compare the values obtained for the pressure of 3.00 mol CO2 at 298.15 K held in a 8.25-dm3 bulb using the ideal gas, van der Waals, Dieterici, and Beattie-Bridgeman equations. For CO2 the Dieterici equation constants are a = 0.462 Pa m6 mol–2, b = 4.63 × 10–5 m3 mol–1 Solution *1.64. A gas obeys the van der Waals equation with Pc = 3.040 × 106 Pa (= 30 atm) and Tc = 473 K. Calculate the value of the van der Waals constant b for this gas. Solution *1.65. Expand the Dieterici equation in powers of Vm1 in order to cast it into the virial form. Find the second and third virial coefficients. Then show that at low densities the Dieterici and van der Waals equations give essentially the same result for P. Solution

1-12

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Essay Questions

Essay Questions 1.66.

In light of the van der Waals equation, explain the liquefaction of gases.

1.67. State the postulates of the kinetic molecular theory of gases. 1.68. Eq. 1.22 defines the ideal-gas thermometer. Describe how an actual measurement would be made using such a thermometer starting with a fixed quantity of gas at a pressure of 150 Torr.

1-13

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Solutions Calculate the amount of work required to accelerate a 1000-kg car (typical of a Honda Civic) to 88 km hr–1 (55 miles hr–1). Compare this value to the amount of work required for a 1600-kg car (typical of a Ford Taurus) under the same conditions.

1.1.

Solution: Given: Car 1 (Civic): m  1000 kg, Speed  88 km hr –1 Car 2 (Taurus): m  1600 kg, Speed  88 km hr –1 Required: work required for the acceleration of each vehicle Any type of work can be resolved through dimensional analysis as the application of a force through a distance; l

w   F (l ) dl lo

1 Recall that bodies in motion possess kinetic energy defined by; Ek  mu 2 where u is the velocity of the moving body and m is its mass. It 2 is possible to determine the amount of work required for the acceleration of a moving body by applying Newton’s Second Law to the work integral given above. l

t

l0

t0

w   F (l )  dl   F (l )  Substitute; F  ma  m t

w  m t0

t dl dt   F (l )  u dt t0 dt

du dt

u l du 1 1  u dt  m  u  du  w   F (l )  dl  mu12  mu02  Ek1  Ek0 l0 u0 dt 2 2

Conversion of speed from km hr–1 to m s-1: Speed = 88 km hr–1  88

km 1 h m  103  24.4 m s -1  s 3600 km h

1-14

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Using the equation for work derived from Newton’s 2nd Law (Civic): l 1 1 . wCivic   F (l )  dl  mu12  mu02  Ek1  Ek0 l0 2 2

1 1 wCivic  (1000 kg)(24.4 m s -1 ) 2  (1000 kg)(0 m s -1 ) 2 2 2 wCivic  297 680 J

wCivic  298 kJ The same method can be applied to the second vehicle (Taurus): l 1 1 wTaurus   F (l )  dl  mu12  mu02  Ek1  Ek0 l0 2 2

1 1 wTaurus  (1600 kg)(24.4 m s -1 ) 2  (1600 kg)(0 m s -1 ) 2 2 2 wTaurus  476 288 J wTaurus  476 kJ By comparing both values quantitatively, it is possible to see that the work required to accelerate a moving body is directly proportional to its mass.

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1-15

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.2.

Solutions

Assume that a rod of copper is used to determine the temperature of some system. The rod’s length at 0 °C is 27.5 cm, and at the temperature of the system it is 28.1 cm. What is the temperature of the system? The linear expansion of copper is given by an equation of the form lt = l0(1 + αt + βt2) where α = 0.160 × 10–4 K–1, β = 0.10 × 10–7 K–2, l0 is the length at 0 °C, and lt is the length at t °C.

Solution:

Given: Copper Rod: l  27.5cm, T  0 C Copper Rod in System: l  28.1cm Linear expansion of copper: lt  l0 (1   t   t 2 ) where α = 0.160×10–4 K–1, β = 0.10×10–7 K–2, l0 is the length at 0 °C, and lt is the length at t °C Required: temperature of the system when the rod length equals 28.1cm Let us define the temperature as t u and make all of the appropriate substitutions into the equation for the linear expansion of copper (starting temperature at zero degrees): lt  l0 (1   t   t 2 )

28.1  27.5(1  0.160 104 t u  0.100 107 t u 2 ) Simplify and rearrange:

28.1 27.5 (1  0.160 104 t u  0.100 107 t u 2 )  27.5 27.5 1.0218  1  1  0.160  10 4 t u  0.100  10 7 t u 2  1

0.0218  0.160  104 t u  0.100  107 t u 2 0  0.100  107 t u 2  0.160 104 t u  0.0218

This can be rearranged to: 0.100 107 x 2  0.160  104 x  0.0218  0 1-16

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Where x  t u and the system can be solved using the quadratic equation:

b  b 2  4ac x 2a x

0.160 104 

 0.160 10   4  0.100 10   0.0218 2  0.100 10  4 2

7

7

x  879o C t u  879o C

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1-17

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.3.

Solutions

Atoms can transfer kinetic energy in a collision. If an atom has a mass of 1 × 10–24 g and travels with a velocity of 500 m s–1, what is the maximum kinetic energy that can be transferred from the moving atom in a head-on elastic collision to the stationary atom of mass 1 × 10–23 g?

Solution:

Given: Atom 1: m1  1 1024 g , u1  500 m s -1

Atom 2: m2  11023 g , u2  0 m s -1

Required: Find Ek(max) that can be transferred from atom 1 to atom 2 It is important to note that during elastic collisions, no energy is lost to the internal motion of the bodies involved. This means that the sums of the kinetic energy in addition to the sums of momentum remain the same before and after the collision. Therefore, there is no potential energy change of interaction between the bodies in collision. Momentum: p  mu 1 Kinetic Energy= Ek  mu 2 2 Conservation of Momentum: m1u1  m2u2  m1u1'  m2u2' Conservation of Energy:

1 1 1 1 m1u12  m2u22  m1u12'  m2u22' 2 2 2 2

(1) (2)

Since u2  0 m s 1 , then we can simplify equation (1) to get: m1u1  m2u2  m1u1'  m2u2'

Rearrangement then gives: u1  u1' 

m2u2' m u'  u1'  u1  2 2 m1 m1

It is possible to substitute the above into equation (2) and solve for u2' ; 1-18

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

2u1 2(500 m s 1 ) ' u   u2  m 1 1023 g 1 2 1 m1 1 1024 g ' 2

u2'  90.9 m s 1 Now this value can be used to find the kinetic energy of atom 2 after the collision. Remember to use SI units by converting grams to kilograms; Ek 

1 m2u2'2 2

1 Ek  (1 1026 kg)(90.9 m s 1 ) 2 2 Ek  4.13  1023 J

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1-19

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.4.

Solutions

Power is defined as the rate at which work is done. The unit of power is the watt (W = 1 J s–1). What is the power that a man can expend if all his food consumption of 8000 kJ a day (≈ 2000 kcal) is his only source of energy and it is used entirely for work?

Solution:

Given: Daily food consumption= 8000 kJ (≈ 2000 kcal) Required: Pone day Remember that power is defined as the rate at which work can be done meaning that; P

dw dt

Since the man’s entire caloric intake is going toward work, then we can say that;

dw  8000 kJ  8000 103 J We are only considering the power exerted in a single day;

dt  1 day  24 hrs  60 min   60 s  dt  24 hrs     86 400 s  1 hr   1 min 

Power is measured by the Watt unit and 1 Watt = 1J s-1 (remember SI units!) P

8000 103 J  92.59 J s 1 86 400 s

P  92.6 W

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1-20

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.5.

Solutions

State whether the following properties are intensive or extensive: (a) mass; (b) density; (c) temperature; (d) gravitational field.

Solution:

Given: (a) mass (b) density (c) temperature (d) gravitational field Required: intensive or extensive? It is first important to define the terms intensive and extensive in the context of physical chemistry. Intensive properties (sometimes called ‘bulk property’) are considered to be physical properties of a system that do not depend on its size. This means that their value will not change when the quantity of the matter in the system becomes subdivided. Extensive properties are the physical properties of a system that DO depend on its size and content. The values of extensive properties change with system subdivision. In addition, the ratio of two intensive properties yields an extensive one. Now it is possible to classify the above properties:

Mass  is extensive as it is a measure of ‘how much’ is present in the system Density  is intensive **note: mass and volume are extensive Temperature  is intensive Gravitational Field  is intensive Back to Problem 1.5

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1-21

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.6.

Solutions

The mercury level in the left arm of the J-shaped tube in Fig. 1.6a is attached to a thermostat gas-containing bulb. The left arm is 10.83 cm and the right arm is 34.71 cm above the bottom of the manometer. If the barometric pressure reads 738.4 Torr, what is the pressure of the gas? Assume that temperature-induced changes in the reading of the barometer and J tube are small enough to neglect.

Solution:

Given: left arm = 10.83 cm, right arm = 34.71 cm, barometric pressure = 738.4 Torr Required: Pgas First, we need to find the difference in heights between the two columns (left and right arms); Right arm - Left arm = 34.71 cm -10.83 cm = 23.88 cm It is important to note that since the arm is open to the atmosphere, this pressure must also be added to the barometric pressure. 1 mmHg = 1 Torr and therefore 23.88 cmHg = 238.8 Torr The pressure of the gas is then found to be; 238.8 Torr + 738.4 Torr = 977.2 Torr Pgas  977.2 Torr

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1-22

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.7.

Solutions

Vacuum technology has become increasingly more important in many scientific and industrial applications. The unit Torr, defined as 1/760 atm, is commonly used in the measurement of low pressures. a. Find the relation between the older unit mmHg and the Torr. The density of mercury is 13.5951 g cm–3 at 0.0 °C. The standard acceleration of gravity is defined as 9.806 65 m s–2. b. Calculate at 298.15 K the number of molecules present in 1.00 m3 at 1.00 × 10–6 Torr and at 1.00 × 10–15 Torr (approximately the best vacuum obtainable).

Solution:

Given: Mercury:   13.5951 g cm –3 , T  0.0 C acceleration of gravity  9.806 65 m s 2

Required: (a) State the relationship between mmHg and Torr (b) NA in V = 1.00 m3 a) We should first define the system as a column of mercury with a 1m2 cross-sectional area, 0.001 m in height, a volume of 0.001 m3. Since we already have the density of mercury it is possible to determine the mass;



m  m  V V

m  13.5951 kg m -3  0.001 m3

Now for 1 mmHg in a column; 1 mmHg   mass  density  acceleration of gravity 



1 mmHg  0.001 m3

 13.5951 kg m  9.806 65 m s  -3

-2

1 mmHg  0.1333 kg m s -2 Now since 1 Torr = 1 mmHg and 1 Torr = 133.322 Pa then we can see that;

1-23

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1 mmHg  133.322 387 4 Pa By definition, 1 atmosphere = 101 325 Pa and 1 Torr = 1/760 atm then; 1 Torr 

1 (101 325 Pa) =133.322 368 4 Pa 760

Therefore; 1 mmHg 

133.322 387 4  1.000 000 14 Torr 133.322 368 4

The Torr is now defined as 1 mmHg. b) Calculate the number of molecules present in a volume of 1.00 m3:

T = 298.15 K, P1 = 1.00×10–6 Torr and P2 = 1.00×10–15 Torr

Using the ideal gas law: PV  nRT we define n as n 

N and rearrange to get; L

PV  nRT

PV 

NRT L

where L is Avogadro’s number and N is the number of particles

L = 6.022  1023 mol-1

And the number density is defined as

N PL  V RT

Remember to make the conversion for pressure! P1 = 1.00×10–6 Torr; N1 P1 L  V RT

1-24

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

 1atm  Torr   760 Torr  N1    101 325 Pa atm 1  1  1 V 8.3145 J K mol 298.15 K

110



6



 



  6.022 10

Solutions

23

mol1



mol1



N1  3.24 1016 m 3 V



N1  3.24  1016 m 3

 1.00 m   3.24 10 3

16

particles

N1  3.24 1016 P2 = 1.00×10–15 Torr using the same method as outlined above; N 2 P2 L  V RT  1atm  Torr   760 Torr  N2    101 325 Pa atm 1  1  1 V 8.3145 J K mol 298.15 K

110



15





N 2  3.24  107 m 3

 



  6.022 10

23

 1.00 m   3.24 10 particles 3

7

N 2  3.24 107 This is still a substantial number!

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1-25

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

The standard atmosphere of pressure is the force per unit area exerted by a 760-mm column of mercury, the density of which is 13.595 11 g cm–3 at 0 °C. If the gravitational acceleration is 9.806 65 m s–2, calculate the pressure of 1 atm in kPa.

1.8.

Solution:

Given: Mercury:   13.595 11 g cm –3 , T  0 C, acceleration of gravity  9.806 65 m s –2 Required: Pcolumn (kPa) Let us define the system as a column of mercury with a cross-sectional area of 1 m2, 0.760 m in height and a volume of 0.760 m3. Since we have the density, it is possible to find the mass of mercury occupying the column;



m V



m  V  13 595.1 kg m 3

  0.760 m  3

m  10 332 kg Mass multiplied by the gravitational acceleration produces a force (or weight) F  ma according to Newton’s Law of Motion. The column’s weight on the unit area then gives a pressure;

Pcolumn  (density)(volume)(acceleration of gravity)



Pcolumn  13 595.1 kg m 3

  0.760 m  9.806 65 m s  3

2

Pcolumn  101 325 kg m s 2 Since 1 Pa = 1kg m s-2 then the pressure is 101.325 kPa.

Pcolumn  101.325 kPa

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1-26

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.9.

Solutions

Dibutyl phthalate is often used as a manometer fluid. Its density is 1.047 g cm–3. What is the relationship between 1.000 mm in height of this fluid and the pressure in torr?

Solution:

Given: Dibutyl phthalate:   1.047 g cm –3 Required: The relationship between 1.000 mm of this manometer fluid and pressure (Torr) When two different liquids are being compared at constant volume and temperature, it is important to note that their pressures will be proportional to their densities. Therefore, it is possible to take the ratio of DBP and Hg densities in order to calculate the pressure associated with 1mm of DBP. 1mmDBP l mmHg

 DBP PDBP P 1.047 g cm 3   DBP   Hg PHg PHg 13.595 g cm 3 PDBP  0.077 PHg PDBP  0.077 Torr Thus, 1mm DBP is equivalent to 0.077 Torr using the fact that 1mmHg is equivalent to 1 Torr. We can also state that; 1 Torr  12.98 mm DBP 0.077 Torr mm 1 1 mm DBP  0.077 Torr

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1-27

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.10. The volume of a vacuum manifold used to transfer gases is calibrated using Boyle’s law. A 0.251-dm3 flask at a pressure of 697 Torr is attached, and after system pumpdown, the manifold is at 10.4 mTorr. The stopcock between the manifold and flask is opened and the system reaches an equilibrium pressure of 287 Torr. Assuming isothermal conditions, what is the volume of the manifold? Solution:

Given: V1  0.251 dm3 , P1  697 Torr , Ppumpdown  10.4 mTorr, Peq  287 Torr Required: Vmanifold Since we are working under isothermal conditions, Boyle’s Law will apply. This law describes the product of pressure and volume for a closed system. In a closed system, the temperature and moles are constant, thus; PV 1 1  PV 2 2 PV  P 1 1 pumpdownV2  Peq (V2  V1 )

 697 Torr   0.251 dm3    0.0104 Torr  V2  287 Torr  V2  0.251 dm3  174.947 Torr dm3   0.0104 Torr  V2   287 Torr  V2  72.037 Torr dm3

Now the above can be simplified on both sides to obtain, 102.91 Torr dm3   286.9896 Torr  V2 V2  Vmanifold 

102.91Torr dm3 286.9896 Torr

Vmanifold  0.359 dm3 Back to Problem 1.10

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1-28

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.11.

Solutions

An ideal gas occupies a volume of 0.300 dm3 at a pressure of 1.80 × 105 Pa. What is the new volume of the gas maintained at the same temperature if the pressure is reduced to 1.15 × 105 Pa?

Solution:

Given: Ideal Gas: V1  0.300 dm3 , P1  1.80 105 Pa Required: V2 In this particular situation, Boyle’s Law will apply. This law describes the product of pressure and volume for a closed system. In a closed system, the temperature and moles are constant, thus; PV 1 1  PV 2 2 Simply rearrange for the final volume (V2); V2 

PV 1 1 P2

1.80 10 Pa   0.300 dm   1.15 10 Pa  5

V2

3

5

V2  0.470 dm3

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1-29

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

If the gas in Problem 1.11 were initially at 330 K, what will be the final volume if the temperature were raised to 550 K at constant pressure?

1.12.

Solution:

Given: same gas as in problem 1.11: V1  0.300 dm3 T1  330 K, T2  550 K (constant pressure) Required: V2 In this particular situation, Charles’ Law will apply. This law states that under constant pressure, the volume of an ideal gas will vary proportionately (by the same factor) with changes in temperature, thus; V1 V2  T1 T2

Simply rearrange for the final volume (V2); V2 

V1T2 T1

 0.300 dm   500 K   3

V2

300 K

V2  0.500 dm3

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1-30

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Calculate the concentration in mol dm–3 of an ideal gas at 298.15 K and at (a) 101.325 kPa (1 atm), and (b) 1.00 × 10–4 Pa (= 10–9 atm). In each case, determine the number of molecules in 1.00 dm3.

1.13.

Solution:

Given: Ideal Gas: T  298.15 K, P1  101.325 kPa 1 atm  , P2  1.00 10 –4 Pa 10 –9 atm  Required: C (in mol dm–3) N A (in V = 1.00 dm3) Knowing that concentration is equal to: C

n V

we can make the substitution into the Ideal Gas Law. C

n P  V RT

For pressure (a) and using the fact that m3 = J Pa-1: C1  C1 

n P  1 V RT 1.013 25  105 Pa

8.3145 J K

1



mol1 (298.15 K )

C1  40.87 mol m 3 Now convert units into mol dm-3: C1 

40.87 mol m 3  0.0409 mol dm 3 103

1-31

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

C1  0.0409 mol dm 3 Number of molecules per unit volume;

 0.0409 mol dm   L    0.0409 3

N1 

2.46 1022 molecules dm3 1.00 dm

3



mol dm 3 6.022  1023 molecules mol1



 2.46 1022 molecules

N1  2.46 1022 molecules For pressure (b) using the same method: C2  C2 

n P2  V RT 1.00  104 Pa

8.3145 J K

1



mol1 (298.15 K )

C2  4.03 108 mol m 3 Now convert units into mol dm-3; 4.03 108 mol m 3 C2   4.03 1011 mol dm 3 3 10 C2  4.03 1011 mol dm 3 Number of molecules per unit volume;

 4.03 10

11





mol dm 3   L   4.03 1011 mol dm 3 6.022  1023 molecules mol1

 1-32

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

N2 

2.43 1013 molecules dm3 1.00 dm

3

Solutions

 2.43 1013 molecules

N 2  2.43 1013 molecules

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1-33

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.14. A J-shaped tube is filled with air at 760 Torr and 22 °C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a funnel into the open end. When the mercury spills over the top of the short arm, what is the pressure on the trapped air? Let h be the length of mercury in the long arm. Solution:

Given: J-Tube: P  760 Torr, T  22 C, long arm h  100 cm, short arm l  40 cm Required: P of trapped air The temperature is again held constant (same as in problems 1.10 and 1.11) so Boyle’s Law will apply; PV 1 1  PV 2 2 We are given the initial pressure, so we can rearrange this equation to solve for P2 ; PV P2  1 1 V2 Since h, the height of the mercury column on the trapped air side (long arm) is proportional to the volume of a uniform tube then we can write;

P2 

P1  100 cmHg (100  h) cmHg

where h is the final height in centimeters of mercury in the long arm. In the short arm;

P2  40  h  P1 Substituting this into the above equation in order to eliminate P2 gives; 40  h  P1 

P1 100 cmHg (100  h) cmHg

1-34

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Recall that 1mmHg = 1 Torr and we therefore can make the substitution for P1 ;

P1 (100)  40  h  P1 (100  h) P1 (100)  (100  h)  40  h  P1  This can be expanded to obtain; P1 (100)  4000  100h  100 P 1  40h  h 2  Ph 1

h 2  140h  76h  4000  0 h 2  216h  4000  0 Using the quadratic equation then yields:

h  195.5 cmHg or h  20.5 cmHg The first value of h cannot be this large since the tube length is only 100 cm. Therefore,

h  20.5 cmHg is the correct value. The final pressure can now be found; P (100) 76 cmHg(100 cm ) P2  1   95.6 cmHg (100  h) (100  20.5) cm

P2  956 Torr Back to Problem 1.14

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1-35

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.15. A Dumas experiment to determine molar mass is conducted in which a gas sample’s P, θ, and V are determined. If a 1.08-g sample is held in 0.250 dm3 at 303 K and 101.3 kPa: a. What would the sample’s volume be at 273.15 K, at constant pressure? b. What is the molar mass of the sample? Solution:

Given: m  1.08 g, V  0.250 dm3 , T  303 K, P  101.3 kPa Required: Vsample and M sample Since we are working under constant pressure, Charles’ Law can be applied. This law states that V1 V  constant  2 T1 T2

Solving for V2 , we obtain VT V2  1 2 T1 Remember that the initial temperature is T = 303 K so by making the appropriate substitutions we will have;

 0.250 dm   273.15 K   3

V2

303 K

V2  0.225 dm3

Now that we have the final volume, it is possible to find the molar mass according to the equation; mRT M PV Recall that in order to derive this equation we must start with the ideal gas law;

1-36

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

m so we obtain M m mRT PV  RT  M  M PV 3 1.08 10 kg  (8.3145 J K 1 mol1 )(273.15 K)  M (101.3 103 Pa)(0.225 dm3 )(103 m3 dm 3 ) PV  nRT and n 

M  0.1076 kg mol-1  M  108 g mol1 Back to Problem 1.15

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1-37

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.16. A gas that behaves ideally has a density of 1.92 g dm–3 at 150 kPa and 298 K. What is the molar mass of the sample? Solution:

Given: Ideal Gas:  = 1.92 g dm–3, P = 150 kPa, T = 298 K Required: M sample Starting with the Ideal Gas Law is it possible to make substitutions and rearrangements in order to solve for the molar mass. PV  nRT

P

nRT V

Now, using the fact that n 

m we can make the next substitution; M

m RT M P V

Since density is defined as   P

M

 RT M

 RT P

m then we can write; V

and now solve for M  M 



 RT P

1.92 kg m 3  8.3145 J K 1 mol1  298.15 K 150  103 Pa

M  0.0317 kg mol1 M  31.7 g mol1 Back to Problem 1.16

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1-38

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.17.

Solutions

The density of air at 101.325 kPa and 298.15 K is 1.159 g dm–3. Assuming that air behaves as an ideal gas, calculate its molar mass.

Solution:

Given: Air:   1.159 g dm –3 , T  298.15 K, P  101.325 kPa Required: M air Use the same method as the previous problem (1.16); M

M

 RT P

 RT P



1.159 kg m 3  8.3145 J K 1 mol1  298.15 K 101 325 Pa

M  0.0284 kg mol1 M  28.36 g mol1

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1-39

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.18.

Solutions

A 0.200-dm3 sample of H2 is collected over water at a temperature of 298.15 K and at a pressure of 99.99 kPa. What is the pressure of hydrogen in the dry state at 298.15 K? The vapor pressure of water at 298.15 K is 3.17 kPa.

Solution:

Given: H2 (over water): V  0.200 dm3 , T  298.15 K, Pt  99.99 kPa Vapor pressure of water: 3.17 kPa at T = 298.15 K Required: PH2 in the dry state This problem makes use of Dalton’s Law of Partial Pressures which states: The total pressure observed for a mixture of gases is equal to the sum of the pressure that each individual gas would exert had it been alone occupying the container and at the same temperature. Pi = xiPt

Partial pressure is defined as the total pressure multiplied by the mole fraction of a particular gas in the mixture. For this particular hydrogen/water system, we can then write; Pt  PH 2  PH 2O and solve for the pressure of hydrogen;

Pt  PH 2O  PH 2

PH2  99.99 kPa  3.17 kPa PH2  96.82 kPa

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1-40

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

What are the mole fractions and partial pressures of each gas in a 2.50-L container into which 100.00 g of nitrogen and 100.00 g of carbon dioxide are added at 25 °C? What is the total pressure?

1.19.

Solution:

Given: Container: V  2.50 L, T  25 o C Add 100.00 g of nitrogen and carbon dioxide Required: xi , Pi for each and Pt First find the amount of each gas in terms of moles because we are provided with their mass and can easily find their molar mass; nN2 

100.00 g m  M 28.012 g mol1

nN2  3.5699 mol

nCO2 

100.00 g m  M 44.010 g mol1

nCO2  2.2722 mol Now we can find the mole fractions associated with each gas using the individual and combined number of moles; xN 2 

nN2 ntot



3.5699 mol (3.5699  2.2722) mol

xN 2  0.6111 xCO2 

nCO2 ntot



2.2722 mol (3.5699  2.2722) mol

xCO2  0.3889 1-41

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Knowing the container volume and temperature of the system, the partial pressures can be calculated using the ideal gas law; PN2 

nN2 RT Vcont



 3.5699 mol  8.3145 J K

1

mol1

  298.15 K 

2.50 dm3

PN2  35.4 bar PCO2 

nCO2 RT Vcont



 2.2722 mol  8.3145 J K

1

mol1

  298.15 K 

2.50 dm3

PCO2  22.5 bar The total pressure is now found using Dalton’s Law for Partial Pressures; Pt  P1  P2  P3    Pi Pt  x1 Pt  x2 Pt  x3 Pt    xi Pt Pt 

(Eq. 1.53)

n RT n1 RT n2 RT   i V V V

Pt  (n1  n2    ni )

RT V

(Eq. 1.54)

Any of the above forms can be used but for simplicity, we shall use Eq. 1.54; Pt  (3.5699  2.2722) mol

(8.3145 J K 1 mol1 )(298.15 K ) 2.50 dm3

Pt  57.9 bar Notice, once you take the pressure, you need to divide by 102 in order to get the pressure in units bar. Back to Problem 1.19

Back to Top 1-42

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.20.

Solutions

The decomposition of KClO3 produces 27.8 cm3 of O2 collected over water at 27.5 °C. The vapor pressure of water at this temperature is 27.5 Torr. If the barometer reads 751.4 Torr, find the volume the dry gas would occupy at 25.0 °C and 1.00 bar.

Solution:

Given: KClO3: VO2  27.8 cm3 , T  27.5 °C Vapor pressure of water: P = 27.5 Torr Barometer reading: P = 751.4 Torr Required: Vdry gas First it is possible to find the pressure of the dry gas at T = 27.5 °C by making use of the barometer reading and the vapor pressure of water; Pgas  Pbarometer  Pwater

Pgas  751.4 Torr  27.5 Torr Pgas  723.9 Torr (Remember that this is at 27.5 °C) Since there is also a temperature change the following equality should be used to find the final volume of the system; PV PV 1 1  2 2 T1 T2 V2 

PV 1 1T2 T1 P2

Recall that 1 bar = 750.06 Torr. Also, when making temperature conversions between Celsius to Kelvin: 27.5 °C = 273.15 + 27.5 = 300.65 K. It is important to remember the initial conditions of the system (Don’t mix up the temperatures!); V2 

PV (723.9 Torr )(27.8 cm3 )(298.15 K ) 1 1T2  T1 P2 (300.65 K )(750.06 Torr )

1-43

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

V2  Vdry gas  26.6 cm3

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1-44

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.21. Balloons now are used to move huge trees from their cutting place on mountain slopes to conventional transportation. Calculate the volume of a balloon needed if it is desired to have a lifting force of 1000 kg when the temperature is 290 K at 0.940 atm. The balloon is to be filled with helium. Assume that air is 80 mol % N2 and 20 mol % O2. Ignore the mass of the superstructure and propulsion engines of the balloon. Solution:

Given: Balloon lifting force: m = 1000 kg, T = 290 K, P = 0.940 atm Required: Vballoon The lifting force comes from the difference between the mass of air displaced and the mass of the helium that replaces the air. We can work under the assumption that the molar mass for air is 28.8 g mol-1. This is true if we consider the fact that air (in the problem) is composed 80 percent of nitrogen and 20 percent of oxygen. M N2  14(2)  28 g mol1

But we will only consider 80 percent and therefore; M N2  28 g mol1 (0.80)  22.4 g mol1 (in air) M O2  16(2)  32 g mol1

But we will only consider 20 percent and therefore; M O2  32 g mol 1 (0.20)  6.4 g mol1

Lifting force V ( air   helium )  1000 kg And recall that we can use the ideal gas law to solve for the density of a gas (density is mass divided by volume);

PV  nRT m RT  nRT M P   RT and solve for density V V M

1-45

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases



PM RT

 air

 0.940 atm  101 325 Pa atm   28.8 g mol   8.3145 J K mol   290 K  10 g kg  1

1

Solutions

1

1

1

3

air  1.138 kg m 3

 0.940 atm  101 325 Pa atm   4.003 g mol   8.3145 J K mol   290 K  10 g kg  1

 helium

1

1

1

3

1

 helium  0.158 kg m 3 Now these values can be substituted into the equation for the volume of the balloon; Vballoon 

Vballoon 

1000 kg ( air   helium ) 1000 kg (1.138  0.158) kg m 3

Vballoon  1021 m3

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1-46

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.22. A gas mixture containing 5 mol % butane and 95 mol % argon (such as is used in Geiger-Müller counter tubes) is to be prepared by allowing gaseous butane to fill an evacuated cylinder at 1 atm pressure. The 40.0-dm3 cylinder is then weighed. Calculate the mass of argon that gives the desired composition if the temperature is maintained at 25.0 °C. Calculate the total pressure of the final mixture. The molar mass of argon is 39.9 g mol–1. Solution:

Given: Gas mixture: 5 mol % butane and 95 mol % argon P  1 atm, Vcyl  40.0 dm3 , M argon  39.9 g mol –1

Required: margon and Pt By using the information given above, it is possible to find the mole fractions for each of the gases in the mixture;

PV  nRT n

PV RT

nbutane 

101 325 Pa   40.0 dm3 

8.3145 J K

1



mol1 298.15 K



nbutane  1.63 mol Since the mixture contains 95 parts argon to 5 parts of butane, the ratio is then 95/5 = 19:1 and we can determine the number of moles for argon; nargon  19 nbutane

nargon  19(1.63) mol  30.97 mol Now that we have both the number of moles and molar mass, we can find the mass of argon; margon  nargon M argon

1-47

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases



margon  30.97 mol

Solutions

  39.9 g mol  1

margon  1236.7 g The total pressure can then be found by taking the sum of the partial pressures; Pt  ( nbutane  nargon )

RT V

Remember that once you find the pressure, you must divide by 102 in order to convert to bar.

8.3145 J K P  1.63  30.97  mol t

1

mol1

  298.15 K 

40.0 dm3

Pt  20.2 bar Since 1 bar = 0.986 92 atm, then we can say that; Pt  20.2(0.986 92) Pt  19.9 atm

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1-48

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.23.

Solutions

The gravitational constant g decreases by 0.010 m s–2 km–1 of altitude. a. Modify the barometric equation to take this variation into account. Assume that the temperature remains constant. b. Calculate the pressure of nitrogen at an altitude of 100 km assuming that sea-level pressure is exactly 1 atm and that the temperature of 298.15 K is constant.

Solution:

Given: gravitational constant g decreases by 0.010 m s–2 km–1 of altitude Required: (a) modify

dP  Mg   dz  RT  P

(b) PN2 at z = 100 km, P = 1 atm, T = 298.15 K a) The standard gravitational acceleration is defined as 9.807 m s-2. If g were to decrease by 0.010 m s-2 per each kilometer in height, this would be equivalent to a change of: 0.010 m s 2  105 s 2 z 3 10 m

where z is the altitude. The new gravitational constant expression would be as follows: g  9.807 m s 2  105 s 2 z This can then be substituted into the Barometric Distribution Law equation, dP  Mg   dz  RT  P

(Eq. 1.74)

To give: dP M  9.807 m s 2  105 s 2 z  dz  P RT This can also be expressed in the following manner: 1-49

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

P M ln     9.807 m s 2 z  5  106 s 2 z 2   RT  P0 

b) The second version of this equation can then be used to calculate the pressure of nitrogen gas at an altitude of 100 km.







28.0 g mol1 103 kg g 1 P ln     8.3145 J K 1 mol1 298.15 K  P0 





 9.807 10  m s 5

2

z  5 106 105  s 2 z 2 2



P ln    10.51  P0   P  10.51  2.73 105  e P  0 P  2.73  105  P0  1 atm 

P  2.73 105 atm

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1-50

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.24. Suppose that on another planet where the atmosphere is ammonia that the pressure on the surface, at h = 0, is 400 Torr at 250 K. Calculate the pressure of ammonia at a height of 8000 metres. The planet has the same g value as the earth. Solution:

Given: Planet with ammonia atmosphere: h = 0, P = 400 Torr, T = 250 K Required: PNH3 at h = 8000 m We may begin as we did in the previous problem with the Barometric Distribution Law: dP  Mg   dz  RT  P

(Eq. 1.74)

We can then integrate this expression, with the boundary condition that P = P0 when z = 0, which yields; ln

P Mgz  P0 RT

(Eq. 1.75)

We can further manipulate the equation by exponentiating each side: ln e

P  Mgz  e RT and solve for P P0

P  P0 e



Mgz RT

Assume that the temperature remains constant at T = 250 K and the molar mass of ammonia is M = 0.017 kg mol-1. These values can be substituted into the above equation. P  (400 Torr)e

( 0.017)(9.807)(8000) (8.3145)(250)

P  (400 Torr)e 0.642 P  210 Torr Back to Problem 1.24

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Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.25. Pilots are well aware that in the lower part of the atmosphere the temperature decreases linearly with altitude. This dependency may be written as T = T0 – az, where a is a proportionality constant, z is the altitude, and T0 and T are the temperatures at ground level and at altitude z, respectively. Derive an expression for the barometric equation that takes this into account. Work to a form involving ln (P/P0). Solution:

Given: linear dependency of temperature on altitude: T = T0 – az Required: Derive an expression for the barometric equation taking linearity of temperature increase into account. Beginning with the Barometric Distribution Law equation (Eq. 1.74), and substituting for T from the linear dependency of temperature on altitude equation; dP  Mg   dz  RT  P

dP Mg  dz P R T0  az  This is a differential equation. In order to solve this, let x  T0  az  then we have:

dx  adz and rearrangement gives  dz 

 dx a

Integration of the expression is then as follows (and with the proper substitutions):

dx 1  x   ln   xo ax a  x0 



x

dx 1  T0  xo ax  a ln  T0  az  x

1-52

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Integration of the LHS between the values P0 and P (with the final substitution) gives:

 P  Mg  T0  az  ln    ln    P0  Ra  T0 

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1-53

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

An ideal gas thermometer and a mercury thermometer are calibrated at 0 °C and at 100 °C. The thermal expansion coefficient for mercury is:

1.26.



1 ( V / T ) P V0

 1.817  10 4  5.90  109   3.45  1010  2 where θ is the value of the Celsius temperature and V0 = V at θ = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 °C? Solution:

Given: Thermometers: T1  0 C, T2  100 C Thermal expansion coefficient for mercury:  

1 (V / T ) P  1.817 104  5.90 109   3.45 1010  2 V0

Required:  Hg when ideal gas scale reads 50 °C In the case of a mercury column, we assign its length the value l100 when it is at thermal equilibrium with boiling water vapor at 1atm pressure. The achievement of equilibrium with melting ice exposed to 1atm pressure establishes the length, l0 . Assuming a linear relationship between the temperature and the thermometric property (length) we can write;



 l  l0 

 l100  l0 

(100 C)

(Eq. 1.15)

This expression can be tailored to the situation given above by;

 Hg 

V50  V0 Hg (100 C) V100  V0 Hg

Since  

1 (V / T ) P then we can integrate the expression with respect to α to get; V0

1-54

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

50

V50  V0   V0 d 0

which can be evaluated as follows (divide the second and third terms by 2 and 3 respectively) 1.817 104  

50 0

 2.95 109  2

50 0

50  1.15 1010  3  V0 0 

 0.009 107 V0 The same can be done for the denominator in the equation  Hg 

V50  V0 Hg (100 C) V100  V0 Hg

1.817 104  100  2.95 x109  2 100  1.15 1010  3 100  V 0  0 0   0  0.018 31V0 Now these two values can be substituted into the above equation to get;

 Hg 

0.009 107 V0 (100 C) 0.018 31 V0

 Hg  49.7 C

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1-55

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.27. It takes gas A 2.3 times as long to effuse through an orifice as the same amount of nitrogen. What is the molar mass of gas A? Solution:

Given: Gas A: teffusion = 2.3 times longer than nitrogen Required: M Gas A This particular problem makes use of Graham’s Law of Effusion which states that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles: rate(gas 1) t (gas 2)   rate(gas 2) t (gas 1)

 (gas 2) M (gas 2)  M (gas 1)  (gas 1)

Since we can easily determine the molar mass of nitrogen (N2), we can make the appropriate substitutions and solve for the molar mass of Gas A. M nitrogen  28 g mol-1 vA vnitrogen

vA vnitrogen



tnitrogen tA



M (gas 2) M (gas 1)

2 1 1 28 g mol1  1   28 g mol       MA 2.3 MA  2.3  

   

2

2

28 g mol1 28 g mol1  1  M    A   2 MA  2.3   1     2.3 

M A  1.5 102 g mol1 Back to Problem 1.27

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1-56

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.28.

Solutions

Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, takes 5.80 minutes to effuse through an orifice. How long will it take for helium to effuse under the same conditions?

Solution:

Given: Vnitrogen  1 dm3 , P  1 bar, t  5.8 min Required: tHe Using Graham’s Law of effusion (as in previous problem 1.27) recall that effusion time is inversely proportional to the rate of effusion. rate(N 2 ) tHe   rate(He) t N2

M He M N2

Rearrange the above equation to isolate for the wanted variable, tHe ; tHe  t N2

M He M N2

We can determine the molar masses of both helium and nitrogen to get; tHe  5.80 min

4 g mol1 28 g mol1

tHe  2.19 min Back to Problem 1.28

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1-57

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.29. What is the total kinetic energy of 0.50 mol of an ideal monatomic gas confined to 8.0 dm3 at 200 kPa? Solution:

Given: Ideal monatomic gas: n  0.50 mol , V  8.0 dm3 , P  200 kPa Required: Ektot This particular problem refers to the section of Kinetic Theory of Gases. Here, we are trying to determine the relationship between u 2 and T, the mechanical variable of u of Eq. 1.41: P

Nmu 2 3V

which is the fundamental equation derived from the simple kinetic theory of gases. For our purpose of determining this relationship (kinetic energy and temperature), Eq. 1.41 may be converted into another useful form by recognizing that the average kinetic energy per molecule is defined as; 1 Єk  mu 2 2 Substitution of this expression into Eq. 1.41 then gives; PV 

1 2 N 2Єk  N Єk 3 3

At constant pressure, the volume of a gas is proportional to the number of molecules and the average kinetic energy of the molecules. Since N  nL then we can write; 2 nLЄk and since LЄk is the total kinetic energy per mole of gas, then 3 2 PV  nEk 3 3 PV Ek  2 n PV 

1-58

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

The data given above can be substituted into the above equation to yield; Ek 

3 (200 kPa)(8.0 dm3 ) 2 0.5 mol

Ek  4800 J mol1

So for half a mole, the kinetic energy will be: nEk  2400 J

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1-59

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.30. Nitrogen gas is maintained at 152 kPa in a 2.00-dm3 vessel at 298.15 K. If its molar mass is 28.0134 g mol–1 calculate: a. The amount of N2 present. b. The number of molecules present. c. The root-mean-square speed of the molecules. d. The average translational kinetic energy of each molecule. e. The total translational kinetic energy in the system. Solution: Given: Nitrogen: P = 152 kPa, V = 2.00 dm3, T = 298.15 K, M = 28.0134 g mol–1 Required: see above a through e Using the ideal gas law, PV  nRT we can solve for the number of moles present. 152 000 Pa  (2.00 dm3 )(103 m3 dm 3 )  PV n  RT (8.3145 J K 1 mol1 )(298.15 K ) n  0.1226 mol We can now use Avogadro’s number in order to find the number of molecules present; number molecules   number of moles  L  N  nL  0.1226 mol (6.022  1023 mol1 )

N  7.38  1022 We can take the square root of Eq. 1.43 in order to find the root mean square speed of the molecules; 3RT (Eq. 1.43) u2  M 3RT u2  M u2 

3(8.3145 J K mol1 )(298.15 K ) 0.028 013 4 kg mol1

u 2  515.2 m s 1 1-60

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

The average translational energy (for each molecule) is given by Eq. 1.44;





1 1 2 1 1 0.0280134 kg mol (515.2 m s ) 2 Єk  mu  2 2 6.022 1023 mol1

Єk  6.175  1021 J

It is possible to find the total translational kinetic energy in the system by using the equation; 3 Ektot  nRT (Eq. 1.49) 2 3 Ektot  (0.1226 mol )(8.3145 J K 1 mol1 )(298.15 K) 2 Ektot  456 J

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1-61

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.31. By what factor are the root-mean-square speeds changed if a gas is heated from 300 K to 400 K? Solution: Given: T1 = 300 K, T2 = 400 K Required: change in root-mean-square speeds Recall Eq. 1.43 3RT u2  M Remember that in problem 1.29 we outlined the relationship between u 2 and T. Using this information, it is possible to see that the following ratios are equivalent; u22 u12



T2 T1

Now we can determine the magnitude of change in root-mean-square speed when moving from a lower to a higher temperature. T2 400   1.33 T1 300 T2 T1

 1.15

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1-62

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.32. The collision diameter of N2 is 3.74 × 10–10 m at 298.15 K and 101.325 kPa. Its average speed is 474.6 m s–1. Calculate the mean free path, the average number of collisions ZA experienced by one molecule in unit time, and the average number of collisions ZAA per unit volume per unit time for N2. Solution: Given: d A  3.74  10 10 m, T  298.15 K, P  101.325 kPa, uA  474.6 m s -1 Required:  , Z A , Z AA The mean free path is given by Eq. 1.68; V l= 2pd A2 N A   Using the ideal gas law, PV = nRT , and solving for V, a useful expression for the mean free path can be obtained; V=

nRT P   

Giving the mean free path as, nRT P l= 2pd A2 N A L=

l=

NA , where N A is the number of particles n

RT 2pd A2 LP  

1-63

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

8.3145 J K mol   298.15 K  2  3.74  10 m   6.022 10 mol  101 325 Pa  1



Solutions

10

2

1

23

1

  6.537 108 J m 2 Pa 1 where 1 J  kg m 2 s 2 and 1 Pa  1 kg m 1s 2 2 2 1 J 1 kg m s =  1 m3 1 2 1 Pa 1 kg m s

  6.537 108 m 2 m3

 

  6.54  10 8 m The average number of collisions ZA experienced by one molecule in unit time, also known as the collision frequency for one molecule is given by Eq. 1.66; ZA 

2 d A2 u A N A V

(SI unit :s 1 )

Using the ideal gas law PV = nRT and solving for V, a useful expression for ZA can be obtained. V= ZA  L=

nRT   P 2 d A2 u A N A P nRT NA , where N A is the number of particles n

1-64

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Z A  7 259 759 289 m3 s 1 Pa J 1

2 d u A LP RT 2 A

ZA 

where 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2



2  3.74  1010 m   474.6 m s 1  6.022  1023 mol1 2

ZA 

Solutions

8.3145 J K

1

mol1

  298.15 K 

 101 325 Pa  11PaJ  1 kg m

1

s 2

2

2

1 kg m s

 1 m 3

Z A  7.26  109 m3 s 1 m 3

Z A  7.26  109 s 1

The average number of collisions ZAA per unit volume per unit time for N2, also known as the collision density is given by Eq. 1.65; 2 d A2 u A N A2 Z AA  2V 2

(SI unit :m 3 s 1 )

Using the ideal gas law PV = nRT and solving for V, a useful expression for ZAA can be obtained; V=

nRT   P

Z AA 

L=

2 d A2 u A N A2 P 2 2  nRT 

2

NA , where N A is the number of particles n

Z AA 

 d A2 u A L2 P 2 2  RT 

2

Solving for ZAA,

1-65

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

2

Z AA 



  3.74  1010 m   474.6 m s 1  6.022  1023 mol1



2 8.3145 J K 1 mol1

 101 325 Pa 

  298.15 K  2

2

Solutions

2

2

Z AA  8.934 67 1034 m3 s 1 Pa 2 J 2 where 1 J  kg m 2 s 2 and 1 Pa  1 kg m 1s 2 2 4 2 1 Pa 2 1 kg m s  1m 6  2 4 -4 1 J2 1 kg m s

Z AA  8.934 67 1034 m3 s 1 m 6 Z AA  8.93  1034 m -3 s -1

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1-66

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.33. Express the mean free path of a gas in terms of the variables pressure and temperature, which are more easily measured than the volume. Solution: Given: l =

V 2pd A2 N A

Required: mean free path in terms of P and T The mean free path is given by; l=

V 2pd A2 N A

(Eq. 1.68)

Using the ideal gas law PV = nRT and solving for V, a useful expression for the mean free path can be obtained; V=

nRT P

nRT P l= 2pd A2 N A L=

l=

NA , where N A is the number of particles n

RT 2pd A2 LP

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1-67

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.34. Calculate ZA and ZAA for argon at 25 °C and a pressure of 1.00 bar using d = 3.84 × 10–10 m obtained from X-ray crystallographic measurements.

Solution: Given: d A  3.84  1010 m, T  298.15 K , P  105 Pa Required: Z A , Z AA ZA is given by Eq. 1.66; 2 d A2 u A N A ZA  V

(SI unit :s 1 )

Using the ideal gas law PV = nRT and solving for V, a useful expression for ZA can be obtained. V= ZA  L= ZA 

nRT   P 2 d A2 u A N A P nRT NA , where N A is the number of particles n 2 d A2 u A LP RT

To solve for ZA the speed must first be found. The average speed is given in the Key Equations section of the chapter;

1-68

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

u

u

8 RT M



  298.15 K   10 kg g 

Solutions

8 8.3145 J K -1 mol1



 39.948 g mol1

3

1

u  158 021.4434 J kg 1

 

since 1 J  1 kg m 2 s 2





u  158 021.4434 kg m 2 s 2 kg 1 u  397.519 m s 1

Solving for ZA gives;

2 d A2 u A LP ZA  RT



2  3.84 1010 m   397.519 m s 1  6.022  1023 mol1 2

ZA 

8.3145 J K

1

mol1

  298.15 K 

 10 000 Pa 

Z A  6 326 376 149 m3 s 1 Pa J 1 where 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1s 2 1 2 1 Pa 1 kg m s  1 m 3  2 2 1J 1 kg m s

Z A  6.33 109 m3 s 1 m 3 Z A  6.33  109 s 1

ZAA is given by Eq. 1.65;

1-69

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Z AA 

2 d A2 u A N A2 2V 2

Solutions

(SI unit :m 3 s 1 )

Using the ideal gas law PV = nRT and solving for V, an expression for ZAA is as follows; V=

nRT   P

Z AA 

L=

2 d A2 u A N A2 P 2 2  nRT 

2

NA , where N A is the number of particles n

Z AA 

 d A2 u A L2 P 2 2  RT 

2

Solving for ZAA to get; 2

Z AA 



  3.84×1010 m   397.519 m s 1  6.022×1023 mol-1



2 8.3145 J K 1 mol1

  298.15 K  2

 10 000 Pa  2

2

2

Z AA  7.684 13  1034 m3 s 1 Pa 2 J 2

since 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2 2 4 2 1 Pa 2 1 kg m s   1 m 6 2 4 4 1 J2 1 kg m s

Z AA  7.684 13  1034 m3 s 1 m 6 Z AA  7.68  1034 m 3 s 1

Back to Problem 1.34

Back to Top 1-70

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.35.

Solutions

Calculate the mean free path of Ar at 20 °C and 1.00 bar. The collision diameter d = 3.84 × 10–10 m.

Solution:

Given: T  20 °C  293.15 K, P  1.00 bar  105 Pa, d A  3.84 10 10 m Required: λ The mean free path is given by Eq. 1.68; V l= 2pd A2 N A Using the ideal gas law PV = nRT and solving for V, a useful expression for the mean free path can be obtained. V=

nRT P

nRT P l= 2pd A2 N A L=

l=

NA , where N A is the number of particles n

RT 2pd A2 LP

1-71

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

8.3145 J K mol   293.15 K  2  3.84 10 m   6.022 10 mol  10 000 Pa  -1



Solutions

10

2

1

23

-1

  6.1781 108 J m 2 Pa 1 since 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2 1 kg m 2 s 2 1J  1 m3  1 2 1 Pa 1 kg m s

  6.1781 108 m 2 m3   6.18  10 8 m

Back to Problem 1.35

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1-72

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.36.

Solutions

Hydrogen gas has a molecular collision diameter of 0.258 nm. Calculate the mean free path of hydrogen at 298.15 K and (a) 133.32 Pa, (b) 101.325 k Pa, and (c) 1.0 × 108 Pa.

Solution:

Given: d A  0.258 nm  2.58  1010 m, T  298.15 K Required: λ The mean free path is given by Eq. 1.68; V l= 2pd A2 N A Using the ideal gas law PV = nRT and solving for V, a useful expression for the mean free path can be obtained. V=

nRT P

nRT P l= 2pd A2 N A L=

l=

NA , where N A is the number of particles n

RT 2pd A2 LP

Now using the fact that P = 133.32 Pa we can make the appropriate substitutions to get;

1-73

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

8.3145 J K mol   298.15 K  2  2.58  10 m   6.022 10 mol  133.32 Pa  1



Solutions

10

1

2

23

1

  1.044  104 J m 2 Pa 1 where 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2 1 kg m 2 s 2 1J   1 m3 1 2 1 Pa 1 kg m s

  1.044  104 m 2 m3   1.044  10 4 m With the next pressure (P = 101.325 kPa) we can use the same method as outlined above;

8.3145 J K mol   298.15 K  2  2.58  10 m   6.022  10 mol  101 325 Pa  1



10

2

1

23

1

  1.37 107 J m 2 Pa 1 where 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2 1 kg m 2 s 2 1J   1 m3 1 2 1 Pa 1 kg m s

  1.37 107 m 2 m3

  1.37 107 m For the final pressure P  1.0  108 Pa ;

1-74

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

8.3145 J K mol   298.15 K  2  2.58  10 m   6.022  10 mol  1.0 10 Pa  -1



Solutions

10

2

1

23

1

8

  1.39 1010 J m 2 Pa 1 where 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2 1 kg m 2 s 2 1J   1 m3 1 2 1 Pa 1 kg m s

  1.39 1010 m 2 m3   1.39  10 10 m Back to Problem 1.36

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1-75

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.37. In interstellar space it is estimated that atomic hydrogen exists at a concentration of one particle per cubic meter. If the collision diameter is 2.5 × 10–10 m, calculate the mean free path λ. The temperature of interstellar space is 2.7 K. Solution:

Given: d A  2.50  10 10 m, T  2.7 K, C  1 particle m 3 Required: λ The mean free path is given by Eq. 1.68; V l= 2pd A2 N A Concentration is given by;

C

NA , where N A is the number of particles V

Now it is possible to solve for λ

 

1 2 d A2C 1

2  2.50 1010 m  1 particle m 3  2

  3.60 1018 m   3.60  1018 m This is about a hundred times greater than the distance between the earth and the nearest star (Proxima Centauri)!

Back to Problem 1.37

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1-76

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.38. Calculate the value of Avogadro’s constant from a study made by Perrin [Ann. Chem. Phys., 18, 1(1909)] in which he measured as a function of height the distribution of bright yellow colloidal gamboge (a gum resin) particles suspended in water. Some data at 15 °C are:

height, z/10–6 N, relative number of gamboge particles at height z

5

35

100

47

–3

ρgamboge = 1.206 g cm

ρwater = 0.999 g cm-3 radius of gamboge particles, r = 0.212 × 10–6 m (Hint: Consider the particles to be gas molecules in a column of air and that the number of particles is proportional to the pressure.) Solution:

Given: see above Required: Avogadro’s number, L Since we consider the gamboges particles to be proportional to the pressure, we can write; dP  Mg   dz  RT  P

(Eq. 1.74)

Here, g is the acceleration due to gravity. Taking the integral of both sides and simplifying gives; ln

N Mg  z where M  mL N0 RT

This can then be substituted into the above equation to get; ln

N mLg  z N0 RT

Solving for L; L 

RT N ln mg z N 0

1-77

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Density is given by  

Solutions

m V

Rearranging for the mass of the gamboges particle then gives; m  V

Where V is the volume of the gamboges particle. Since we know that volume is given by; V 

4 r 3 3

We can then define mass as follows; m

4 r 3 3

Now Avogadro’s number can be expressed as; L 

RT  4 3   r  g z   3 

ln

N N0

Solving for Avogadro’s number, L 

8.3145 J K

1



mol1 288.15 K

 4  kg m –3  (1.206 g cm –3  0.999 g cm –3 )  103   3 g cm –3   1 100   ln –6 –6  35 10 m  5 10 m  47



  0.212 10 –6 m  





3

   9.81 m s 2  L  7.439 74  10 23 mol 1  

L  7.44  10 23 mol 1

Back to Problem 1.38

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1-78

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.39. Refer to Table 1.3 (p. 32) and write expressions and values for (a) the ratio u 2 / u , and (b) the ratio ū/ump. Note that these ratios are independent of the mass and the temperature. How do the differences between them depend on these quantities? Solution:

Given: Table 1.3 Required:

u 2 / u and ū / ump

From Table 1.3 the root mean speed is u 2 

3kBT 8kBT , and the average speed is u  m m

u2 3kBT 8kBT   m m u 3 kBT  m u2   m 8 kBT u u2 3  8 u u2  1.085 u From Table 1.3 the average speed is u 

8kBT 2kBT and the most probable speed is ump  m m

1-79

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

ū 8kBT 2kBT   ump m m 8 kBT ū m   ump  m 2 kBT ū 4  ump  ū 2  ump 

ū  1.128 ump The differences between u 2 and u and between ū and ump increase with T and decrease with m.

Back to Problem 1.39

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1-80

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.40. The speed that a body of any mass must have to escape from the earth is 1.07 × 104 m s–1. At what temperature would the average speed of (a) a H2 molecule, and (b) an O2 molecule be equal to this escape speed? Solution: Given: u  1.07  10 4 m s -1 Required: TH2 TO2 , Average speed, as listed in Table 1.3, is given by u 

8kBT m

By rearranging this equation, temperature can be described as; T

 mu

2

8kB

The mass is given by m  2

T

u M 8kB L

where, kB 

M and by using this expression, the temperature can be simplified to; L

R L

2

T

u M 8R

a. Solving for TH 2

;

1-81

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases





 1.07  104 m s 1  2 1.00794 g mol1 103 kg g 1 2

TH2 

Solutions



8 8.3145 J K 1 mol1



TH2  10 898 m 2 s 2 kg J 1 K where 1 J  1 kg m 2 s 2





TH2  10 898 m 2 s 2 kg kg 1 m 2 s 2 K TH2  10 898 K

TH2  1.09 104 K b. Solving for TO2

;

TO2 





 1.07 10 m s-1  2 15.9994 g mol-1 103 kg g -1 2

4



8 8.3145 J K -1 mol-1



TO2  172 992 m 2 s -2 kg J -1 K where 1 J  1 kg m 2 s 2





TO2  172 992 m 2 s -2 kg kg 1 m 2 s 2 K TO2  172 992 K

TO2  1.73 105 K Back to Problem 1.40

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1-82

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.41. a. For H2 gas at 25 °C, calculate the ratio of the fraction of molecules that have a speed 2u to the fraction that have the average speed ū. How does this ratio depend on the mass of the molecules and the temperature? b. Calculate the ratio of the fraction of the molecules that have the average speed ū100 ºC at 100 °C to the fraction that have the average speed ū25 ºC at 25 °C. How does this ratio depend on the mass? Solution: Given: T  25 C  298.15K Required:

a.

dN u1

dN u 2

N u1

Nu 2

b.

dN u1

dN u 2

N u1

Nu 2

, where u1  2u , u 2  u , u 

8kBT m

where u1  u100  C , u 2  u 25  C , u 

8kBT m

a) The key words in this problem are ratio of the fractions, therefore we use the Boltzmann distribution. The Boltzmann distribution is  m  dN given by Eq. 1.91;  4   N  2 k BT 

3/2

e  mu

2

/2 kBT

u 2 du

Solving for the ratio;

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2



 m  4    2 kBT   m  4    2 kBT 

e



2

2



 m u1 u 2 /2 kBT

3/2 2

2

e  mu1 / 2 kBT u1 du

3/2

e  mu 2  u 12  2  u2 

2

/2 kBT

2

u 2 du

   

1-83

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

where u1  2u , u 2  u dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

since u 

e

e

 

2 2  m 2 u  u  /2 kBT  



2

2



 m 4 u u /2 kBT

 

 2u   u 2 

 4 u2   u2 

2

   

   

  2

 4e

 m 3u /2 kBT

8kBT , we can substitute for the average speed; m

dN u1

dN u 2

N u1

Nu 2

dN u1

dN u 2

N u1

Nu 2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

 4e

 4e  4e

  8 k T 2  B  m 3  / 2k T    m   B  

  24 kBT  m    m 



  1     2 kBT  

   

12



 0.087 735 885

 8.77 102

It is now possible to see that the ratio is independent of mass and temperature of the molecules.

1-84

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

b)

dN u1

dN u 2

N u1

Nu2



 m  4    2 kB 

3/ 2

 m  4    2 kB 

3/2

 1     T100 C   1     T25 C 

3/2

e

2

 mu1 /2 kBT100  C

2

u1 du

3/2

e

2

 mu 2 /2 kBT25  C

2

u 2 du

3/ 2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

where u 

 1  2 2   u 2   m  u1   u12  T  k T T 2 100 C  B  100  C 25  C      2 e  3/2  u2   1       T25 C   T100 C    T25 C   

3/2

e

2 2 u 2   m  u1   2 kB T100  C T25  C   

 u12   2  u2   

8kBT m

dN u1

dN u 2

N u1

Nu2

 T100 C    T25 C   

3/ 2

e

  8k T B 100  C   m  m   2 kB  T100  C   

2

  8 kBT25  C     m   T25  C

   

2

       

  8k T B 100  C   m    8kBT25 C   m 

2

    2    

       

1-85

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

 T100 C    T25 C   

3/2

 T100 C    T25 C   

1/2

 T100 C    T25 C   

1/2

 T100 C    T25 C   

1/2

  8 kBT100  C   8 kBT25  C         m    m   m     T100  C T25  C 2 kB      

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

dN u1

dN u 2

N u1

Nu2

 373.15 K     298.15 K 

dN u1

dN u 2

N u1

Nu2

 0.893 872 7

dN u1

dN u 2

N u1

Nu2

e

e

e

 1 m    2 kB 

 8 kB     m 

  T100  C     T100  C

Solutions

 T100  C     T25 C 

  T25  C     T25  C  

   

 4    0   

1/ 2

 0.894

Back to Problem 1.41

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1-86

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.42. Suppose that two ideal gases are heated to different temperatures such that their pressures and vapor densities are the same. What is the relationship between their average molecular speeds? Solution: Given: Two Ideal Gases,  , P and T Required: the relationship between the average speeds of two ideal gases To solve this problem, we use the Ideal Gas Law to eliminate the temperature dependence from the equation for average speed. This is true because for T1 and T2 , P1  P2  P and 1   2   ; PV  nRT PV T nR where R  kB L T

PV nkB L

Substituting the above expression into the equation for average speed, as given in Table 1.3, and simplifying, gives the relationship between the average speed of two ideal gases that are heated to different temperatures such that their pressures and vapor densities are the same. u u

8kBT m 8 kB PV

 mn kB L

8PV  mnL N nmL where    V V 8P u u



1-87

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Since P and ρ are the same, the average speed is the same for both gases.

Back to Problem 1.42

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1-88

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.43. a. If ū25 ºC is the average speed of the molecules in a gas at 25 °C, calculate the ratio of the fraction that will have the speed ū25 ºC at 100 °C to the fraction that will have the same speed at 25 °C. b. Repeat this calculation for a speed of 10 ū25 ºC. Solution: Given: T100 C  373.15 K, T25 C  298.15 K Required: a)

b)

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

The key words in this problem are ratio of the fractions, therefore we use the Boltzmann distribution. The Boltzmann distribution is given  m  dN by Eq. 1.91;  4   N  2 kBT 

3/2

e  mu

2

/2 kBT

u 2 du , where the average speed of molecules is given in Table 1.3 as u 

8kBT m

Solving for the ratio we get;

dNT100  C

dNT  C

NT100  C

NT  C



 m  4    2 kB 

3/2

 m  4    2 kB 

3/ 2

 1     T100 C   1     T25 C 

3/2

e

3/2

e



 m u 25  C



2

/ 2 kBT100  C

u  100  C



 m u 25  C



2

/2 kBT25  C

u  25  C

2

2

du

du

3/ 2

dNT100  C

dNT  C

NT100  C

NT  C

 1  2 2   m   u 25  C   u 25  C       T25  C  u 100  C 2 kB  T100  C T100 C      e 3/ 2  1  u 25  C    T25 C 

 

 

2 2

1-89

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

where u 

Solutions

8kBT m

dNT100  C

dNT  C

NT100  C

NT  C

 T100  C    T25 C   

3/2

e

 T100  C    T25 C   

3/2

 T100  C    T25 C   

1/2

 T100  C    T25 C   

1/2

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

e

e

e

  8k T B 25  C   m  m   2 kB  T100  C   

2

  8 kBT25  C     m   T25  C

  8 kBT25  C   8 kBT25  C        m    m    m     2 kB  T100  C T25  C     

 1 m    2 kB 

 8 kB     m 

   

2

       

2

 8kBT  C       m   2  8kBT25 C      m  

 T100 C     T25  C 

 T25  C T25  C    T100  C T25  C

   

 4  T25  C  1      T100  C 

 373.15 K     298.15 K 

1/ 2

 4  298.15 K  1     373.15 K 

e

 1.154 559

 1.155

At a speed of 10 ū25 ºC;

1-90

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

dNT100  C

dNT  C

NT100  C

NT  C



 m  4    2 kB 

3/2

 m  4    2 kB 

3/2

 1     T100 C 

3/2

 1     T25 C 

e



 m 10 u 25  C

3/2

e



2

/2 kBT100  C

u  100  C



 m u 25  C



2

/2 kBT25  C

u  25  C

2

2

Solutions

du

du

3/2

dNT100  C

dNT  C

NT100  C

NT  C

where u 

 1  2 2   m  10 u 25  C   u 25  C      2 kB  T100  C T25  C  u 100  C T100 C    e   3/2  1  u 25  C    T25 C 

 

 

2 2

8kBT m

1-91

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

dNT100  C

dNT  C

NT100  C

NT  C

 T100  C    T25 C   

3/2

e

 T100  C    T25 C   

3/2

 T100  C    T25 C   

1/2

 T100  C    T25 C   

1/2

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

dNT100  C

dNT  C

NT100  C

NT  C

e

e

e

  10 8 kBT25  C  m  m   2 kB  T100  C   

2

  8 kBT25  C     m   T25  C

  8 kBT25  C   8 kBT25  C  100   m    m   m 2 kB  T100  C T25  C   

 1 m    2 kB 

 8 kB     m 

       

   

2

       

Solutions

2

 8kBT C      m    2  8kBT25  C       m  

 T100  C     T25 C 

 100 T25  C T25  C     T25  C   T100  C

 4  100 T25  C  1       T100  C 

 373.15 K     298.15 K 

1/2

e

 4  100298.15 K  1      373.15 K 

 2.099 87 1044

 2.10 1044

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1-92

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.44. On the basis of Eq. 1.80 with β = 1/kBT, derive an expression for the fraction of molecules in a one-dimensional gas having speeds between ux and ux + dux. What is the most probable speed? Solution: 2

Given: β = 1/kBT, dPx = Be  mux  / 2 du x Required:

dN , the fraction of molecules in a one-dimensional gas N

Using Eq. 1.80, the fraction of molecules in a one-dimensional gas having speeds between ux and ux + dux can be written as; 2

dN dPx Be mux  /2 du x   N P   mux2  /2 du x  Be 0

Using 

 Be 0





0

e

 ax 2

1 dx    2  a

1/ 2

from the appendix in Chapter 1, the denominator can be simplified. 1/2

 mu x2  /2

B  2  du x    2  m 

1-93

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

2

dN Be  mux  /2 du x = 1/2 N B  2  2  m  2

dN 2 Be  mux  /2  2  =   N B  m 

1/2

du x

1/2

2 dN  m  = 2e  mux  /2   du x N  2  1 where   kBT

1/2

1

 mu x2  m  dN = 2e 2 kBT   du x N  2 kBT 

2

1

 mu x dN 2 k BT =e N

1/2

 2m      kBT 

Back to Problem 1.44

du x

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1-94

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.45. Derive an expression for the fraction of molecules in a one-dimensional gas having energies between  x and  x  d  x .Also, obtain

an expression for the average energy  x . Solution: 2

1

 mu x dN 2 kBT Given: =e N

Required:

1/ 2

 2m      kBT 

du x (from problem 1.44)

dN , the fraction of molecules in a one-dimensional gas N

1 Using Eq. 1.93,  x  mu 2 , the fraction of molecules in a one-dimensional gas having speeds between ux and ux + dux 2 1/2 1 2  mu x dN 2 kBT  2 m  , =e   du x (from problem 1.44), can be converted into the fraction molecules having energies between  x and  x  d  x ; N   k BT 

1-95

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1 2

 x  mu x 2 2 x m

ux 



1

du x 1  2 x  2  2       d x 2  m   m  du x 1  2 x     d x m  m  du x 1  1     d x m  m 





1 2

1 2

 2 x 



1 2

1

1 du x  1  2      2 x  2 d x  m  1 du x    2m x  2 d x 1

du x   2m x  2 d  x 

x

 dN  e kBT N

1/2

 1    d x   kBT  x 

The average energy is given by; 

   0

dN N

(Eq. 1.97)

Tailoring the above equation to this particular situation, we get;

1-96

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

dN x 0 N   x   x    x  e kBT 0  

Solutions



x   x

x  



0

Using





0

e



1/2   1    d  x    kBT  x  



1/2

 1   kBxT 1 2   e  x d x   kBT 



x kBT



0

e

 ax

1/ 2

x

1  dx    2a  a 

1/ 2

1 2

from the appendix in Chapter 1, the expression can be simplified. 1

 x1 2 d  x  kBT  kBT  2 1

1  1 2 1 2 x   k T k T    B B   kBT  2 1  x  kBT 2

Solving for the fraction of energies we get;  2 x  1  m  2 kBT

 m dN e  N

1/ 2

 2m      kBT 

 2m x 

x

 2m 1     d x   kBT 2m x 

x

 1    d x   kBT  x 

 dN  e kBT N  dN  e kBT N



1 2

d x

1/2

1/2

1 2

 x  kBT Back to Problem 1.45

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1-97

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.46. Derive an expression for the fraction of molecules in a two-dimensional gas having speeds between u and u + du. (Hint: Proceed by analogy with the derivation of Eq. 1.91.) Then obtain the expression for the fraction having energies between and   d  . What fraction will have energies in excess of  *? Solution:  m  dN Given: Two dimensional gas:  4  N  2 k BT 

Required:

3/ 2

e  mu

2

/ 2 k BT

u 2 du

dN , the fraction of molecules in a two-dimensional gas N

The following equations: 2

dPx = Be  mux  / 2 du x

dPy = Be

 mu 2y  /2

(Eq. 1.80)

du y

(Eq. 1.81)

can be combined to give an expression reflecting the probability that the two components of speed have values between ux and ux + dux, uy and uy + duy.



2

dPx dPy = Be  mux  / 2 du x dPx dPy  B 2 e





 m u x2  u 2y  /2

  Be

 mu 2y  /2

du y



du x du y

Using polar coordinates, we consider a circular shell of radius u and replace du x du y by 2 udu , and take u 2  u x2  u y2 We can then rewrite dPx dPy  B 2 e





 m u x2  u 2y  /2

2

du x du y as dP  2 B 2 e  mu  /2 udu

Using Eq. 1.91, an expression for the speed can be obtained;

1-98

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

2

dN dP 2 B 2 e  mu  / 2 udu   N P  2  mu 2  /2 udu  2 B e 0

2  mu 2  /2

dN 2 B e udu   2 N 2 B 2  e  mu  /2 udu 0

 mu 2  / 2

dN e udu  N  mu 2  /2 e udu  0

Using





0

2

e  ax x dx 

1 from the appendix to Chapter 1, the denominator can be simplified. 2a



1 2   mu 2  /2 e udu    0 2  m 



e

 mu 2  /2

udu 

0

1 m

2

dN e mu  / 2 udu  1 N m 2 dN  m e mu  /2 udu N 1 where   kBT mu 2

dN m  2 kBT  udu e N kBT

1-99

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

mu 2

dN m  2 kBT udu  e N kBT 1 Using Eq. 1.93,   mu 2 , the fraction of molecules in two dimensional gas having speeds between u and u + du., can be converted into the 2 fraction molecules having energies between  and   d  ;

1 2

  mu 2 1

 2  2 u    m  

1

du 1  2  2  2       d 2  m   m  du 1  2     d m  m  du 1  1     d m  m 





1 2

1 2

 2 

1

du  1  2     2 d  m  1 du    2m  2 d du   2m





1 2







1 2

1 2

d

We can now solve for the fraction of energies;

1-100

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

mu 2

dN m  2 kBT udu  e N kBT 1

 2  1

dN m  m m  2 kBT  2  2  e    2m N kBT  m  

1

dN m  kBT  2  2  e    2m N kBT  m 





1 2





1 2

d

d



dN m  kBT   12    12   e  m   m  d N kBT    

1  kBT dN e d  N kBT 

dN 1  kBT  e d N kBT The fraction of molecules with energy greater than  * can be obtained from the expression 

  N*   e kBT d  * N



 N*  e kBT N





*

*

 N*  e kBT N

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1-101

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.47. In Section 1.13 it was stated that the van der Waals constant b is approximately four times the volume occupied by the molecules themselves. Justify this relationship for a gas composed of spherical molecules. Solution:

The Van der Waals constant b represents the excluded volume occupied by the volume of the colliding molecules. When two molecules collide, the closest they can come to one another is a distance of 2r, therefore the excluded volume per molecule can be represented as a sphere with a radius of no less than 2r. 2r

b can be calculated using the volume of a sphere, taking the radius as 2r.  

4 r 3 where r  2r 3

V

V V

4  2r 

3

3 4  8r 3  3

 8V

b  4V Since we only consider the volume occupied by one molecule of radius r, b=4V.

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1-102

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.48. Draw the van der Waals PV isotherm over the same range of P and V as in Figure 1.21 at 350 K and 450 K for Cl2 using the values in Table 1.4. Solution:

Given: Figure 1.21 T=350K, T=450K Required: draw the Van der Waals isotherms The curves are similar to those in Figure 1.21

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1-103

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.49. Compare the pressures predicted for 0.8 dm3 of Cl2 weighing 17.5 g at 273.15 K using (a) the ideal gas equation and (b) the van der Waals equation. Solution:

Given: V  0.8 dm 3 , mCl2  17.5 g, T  273.15 K Required: PIdeal and Pvdw From the ideal gas equation; PIdeal  PIdeal 

nRT V

mRT m , where n  MV M

Solving for PIdeal yields; PIdeal 

17.5 g  8.3145 J K

1

mol1

 2  35.4527 g mol   0.8 dm 1

3

  273.15 K 

 10-3 m3 dm 3



PIdeal  700 658 J m 3 where 1 J  1 kg m 2 s 2 and 1 Pa  1 kg m 1 s 2 PIdeal  700 658 kg m 2 s 2 m 3 PIdeal  700 658 kg m 1 s 2

PIdeal  700.7 Pa Using the van der waals equation; Pvdw Pvdw 

mRT m  M V  M 

 b 



nRT an 2   V  nb V 2

am 2 m , where n  2 2 M V M

1-104

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Solving for Pvdw with a  0.6579 Pa m6 mol –2 and b  0.0562  103 m3 mol –1 we get; Pvdw 

17.5 g  8.3145 J K

 2  35.4527 g mol  1



 0.6579 Pa m

6

2

mol1



  273.15 K  

 17.5 g  0.8 dm3 103 m3 dm 3   2  35.4527 g mol1 





mol –2 17.5 g

 2  35.4527 g mol   0.8 dm 1

1

3





 0.0562 10

3

3

m mol

–1



   

2

 103 m3 dm 3



2

Pvdw  712 997.84 Pa  62 613.823 3 Pa Pvdw  650 384.016 7 Pa Pvdw  650 kPa

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1-105

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.50. A particular mass of N2 occupies a volume of 1.00 L at –50 °C and 800 bar. Determine the volume occupied by the same mass of N2 at 100 °C and 200 bar using the compressibility factor for N2. At –50 °C and 800 bar it is 1.95; at 100 °C and 200 bar it is 1.10. Compare this value to that obtained from the ideal gas law. Solution:

Given: V1  1.00L, T1  –50 C  223.15 K, P1  800 bar, Z1  1.95 T2  100°C  373.15 K, P2  200 bar, Z 2  1.10 Required: V2 and compare to Videal To determine V2, we can use Eq. 1.98 for a real gas and rearrange for n, the number of moles of N2; PV nRT ZPV n RT Z

The number of moles at V2 is the same as the number of moles at V1 since we know that the same mass is used. n1 

Z1 PV Z PV 1 1 , n2  2 2 2 RT1 RT2

where n1  n2 , Z PV Z1 PV 1 1  2 2 2 R T2 R T1 Solving for V2 to get; V2  V2 

Z1 PV 1 1T2 Z 2 P2T1

1.95 800 bar  1.00 L   223.15 K  1.10   200 bar   373.15 K 

V2  3.7747 L V2  3.77 L 1-106

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

To determine Videal, we can use Eq. 1.98 for an ideal gas, with Z = 1 and rearrange for n, the number of moles of N2; PV ,Z 1 nRT PV n RT PV PV n1  1 1 , n2  2 ideal RT1 RT2 Z

where n1  n2 , PV PV 1 1  2 ideal R T2 R T1 Now solving for Videal; Videal  Videal 

PV 1 1T2 P2T1

800 bar  1.00 L   223.15 K   200 bar   373.15 K 

Videal  6.688 77 L Videal  6.69 L We can now compare V2 and Videal by determining the error on Videal.

error 

Videal V2  100% V2

6.69  3.77 100% 3.77 error  0.7745  100% error 

error  77.5% Back to Problem 1.50

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1-107

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.51. A gas is found to obey the equation of state:

P

RT a  V b V

where a and b are constants not equal to zero. Determine whether this gas has a critical point; if it does, express the critical constants in terms of a and b. If it does not, explain how you determined this and the implications for the statement of the problem. Solution:

Given: P 

RT a  where a and b are constants not equal to zero V b V

Required: critical point in terms of a and b, if it exists  2 P   P  According to Eq. 1.99, a gas has a critical point if    0 and  2   0  V Tc  V Tc

RT a  P   2 0    2  V Tc V  b  V a RT  2 2 V V  b 

1

 2 P  2 RT 2a  3 0  2  3  V Tc V  b  V 2a 2 RT  3 3 V V  b 

 2

 2 P   2  should not exist.  P  If    0 and  2   0 , then 1  V Tc  V Tc

1-108

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

 2 RT   3  V  b     a   RT   V 2   2  V  b  

2a   V 3 

2V 2 2 V  b   V3 V  b 3

2

2 2  V V b 1 1  V V b

This last line

1 1  is only true if b = 0, however b ≠ 0 from the statement of the problem. Therefore the gas does not have a critical V V b

point.

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1-109

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

1.52.

Solutions

Ethylene (C2H4) has a critical pressure of Pc = 61.659 atm and a critical temperature of Tc = 308.6 K. Calculate the molar volume of the gas at T = 97.2 °C and 90.0 atm using Figure 1.22. Compare the value so found with that calculated from the ideal gas equation.

Solution: Given: Ethylene (C2H4): Figure 1.22, Pc  61.659 atm, Tc  308.6 K, P  90.0 atm, T  97.2 °C  370.35K

Required: Vm, and Videal First the reduced temperature and pressure of the gas can be obtained using the following ratios; Tr 

Tr 

308.6 K  1.20 370.35 K

Pr 

90.0 atm  1.46 61.659 atm

T P and Pr  Tc Pc

Using Figure 1.22, the compressibility factor for a gas of Tr = 1.20 and Pr = 1.46 is found to be approximately 0.7. Eq. 1.98 gives the compressibility in terms of molar volume. Rearranging this expression for Vm will allow us to calculate the molar volume.

PVm RT ZRT Vm  P Z

Vm 





0.7 0.08206 atm dm3 K 1 mol1 370.35 K

 90.0 atm 



Vm  0.236 37 dm3 mol1 Vm  0.236 dm3 mol1 The molar volume obtained from the ideal gas equation is given by;

1-110

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

V RT = = Vm n P RT Vm = P

Vm

 0.08206 atm dm 

3



K 1 mol1 370.35 K

 90.0 atm 



Vm  0.337 67 dm3 mol1 Vm  0.338 dm3 mol1 A comparison with the ideal molar volume shows that the real molar volume obtained from the law of corresponding states is much smaller.

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1-111

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.53. Assuming that methane is a perfectly spherical molecule, find the radius of one methane molecule using the value of b listed in Table 1.5. Solution: Given: methane (CH4): Table 1.5 Required: rCH4 Using Table 1.5, b  0.0428  103 m3 mol-1 and b  4Vm , as stated in section 1.13, the volume of methane is treated as a sphere; 4 r 3 3 4 r 3 b   4L 3

VCH4  VCH4

We can divide by L, Avogadro’s number, since we are considering only one molecule of methane.

r

3

b 3  4 L 4



0.0428 103 m3 mol1 3 r3  16 6.022  1023 mol1







r  1.618 77  1010 m r  1.62  1010 m The actual radius, i.e. the C-H distance in CH4 is 1.09 1010 m .

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1-112

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.54. Determine the Boyle temperature in terms of constants for the equation of state:

PVm = RT{1 + 8/57(P/Pc)(Tc/T)[1 – 4(Tc/T)2]} R, Pc, and Tc are constants. Solution: 8 P  Given: PVm  RT 1   57  Pc 

  Tc   T

2   Tc    1 – 4        T   

Required: TB, Boyle temperature

   PV   The Boyle Temperature occurs when the second virial coefficient, B(T) = 0 and the partial derivative   becomes zero as P→0.  P T 2

T  This is fulfilled when 1 – 4  c   0 , therefore; T  2

T  1 – 4 c   0  TB  2

 Tc  1    4  TB  Tc 1  TB 2 TB  2Tc

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1-113

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.55. Establish the relationships between van der Waals parameters a and b and the virial coefficients B and C of Eq. 1.117 by performing the following steps: a. Starting with Eq. 1.101, show that

PVm V a 1  m  . RT Vm  b RT Vm b. Since Vm/(Vm – b) = (1 – b/Vm)–1, and (1 – x)–1 = 1 + x + x2 + …,, expand (1 – b/ Vm)–1 to the quadratic term and substitute into the result of part (a). c. Group terms containing the same power of Vm and compare to Eq. 1.117 for the case n = 1. d. What is the expression for the Boyle temperature in terms of van der Waals parameters? Solution: Starting with equation 1.101 P   Vm PVm  Vm  RT    RT  RT  Vm  b  RT

RT a V  2 , we can multiply by m to get; Vm  b Vm RT

 a  2 V  m

PVm V a 1  m  RT Vm  b RT Vm Since Vm/(Vm – b) = (1 – b/Vm)–1, and (1 – x)–1 = 1 + x + x2 + …, therefore we can write; –1

2

 b  b  b      1 –   1 Vm  Vm  Vm   Using the expression derived above; 2

PVm b  b  a 1  1     RT Vm  Vm  RT Vm 1-114

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Grouping the terms containing the same power of Vm gives;

PVm 1  a  1  b  RT Vm  RT Equation 1.117 is

2

  b         Vm 

PV B(T )n C (T )n 2 D(T )n 4  1     . For the case n = 1, this becomes, nRT V V2 V4

B T  C T  D T  PVm 1   2   RT V V V4 a  Comparing to the expression we obtained in part c, we can see that: B T    b  RT 

 2  , C T   b and D T   0 

The expression for the Boyle temperature in terms of van der Waals parameters is determined when B T   0 ;

 a b  RTB  a b RTB TB 

 0 

a bR

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1-115

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.56. Determine the Boyle temperature of a van der Waals gas in terms of the constants a, b, and R. Solution:

Given: A van der Waals gas Required: TB of the constants a, b, and R The temperature can be obtained by rearranging the van der Waals equation;  an 2   P  2   V  nb   nRT V  

P

(Eq. 1.100)

RT a  2 V b V

We can then multiply through by

V to change the form of the equation; RT

PV  V  RT  V  a    RT  RT  V  b  RT  V 2 PV V a   RT V  b RTV V a Z  V  b RTV

1-116

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Since V/(V – b) = (1 – b/V)–1, and (1 – x)–1 = 1 + x + x2 + …, therefore; –1

2

b b b  1 –   1      V V V   2

b b a 1 Z  1      V V  RT V Grouping the terms containing the same power of V gives, 2

1 a  b Z  1  b       V RT   V   Z  The Boyle Temperature occurs when the second virial coefficient, B(T) = 0 and the partial derivative   becomes zero as P→0, i.e.:  P T  Z  lim   0 P 0 P  T

By changing the variable V into

RT we can get; P

1-117

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

P  a Z  1 b  RT  RT

Solutions

2

  b  2   P    RT 

a 1   Z     b  RT  P T RT   Z  lim   0 P 0 P  T

2

  b    2  P   RT 

a  1   Z  lim    b  0 P 0 P RT   T RT  where B T   0 B TB   0  b

a 0 RTB

b

a RTB

TB 

a  1  b   RTB  RTB 

a Rb

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1-118

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.57. The critical temperature Tc of nitrous oxide (N2O) is 36.5 °C, and its critical pressure Pc is 71.7 atm. Suppose that 1 mol of N2O is compressed to 54.0 atm at 356 K. Calculate the reduced temperature and pressure, and use Figure 1.22, interpolating as necessary, to estimate the volume occupied by 1 mol of the gas at 54.0 atm and 356 K. Solution:

The reduced temperature and pressure of the gas can be obtained using the ratios Tr 

T P and Pr  Tc Pc

Using the values above, we obtain Tr 

356 K  273.15  36.5  K

Tr  1.149 69 Tr  1.15

Pr 

54.0 atm 71.7 atm

Pr  0.753138 Pr  0.753

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1-119

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.58. At what temperature and pressure will H2 be in a corresponding state with CH4 at 500.0 K and 2.00 bar pressure? Given Tc = 33.2 K for H2, 190.6 K for CH4; Pc = 13.0 bar for H2, 46.0 bar for CH4. Solution: 

Given: TCH4  500.0 K, PCH4  2.00 bar, TcH  33.2 K, 2

TcCH 190.6 K, PcH  13.0 bar, PcCH  46.0 bar 4

2

4

Required: TH2 and PH2 In order for hydrogen to be in the corresponding state as methane, they must have the same reduced temperature and reduced pressure. The T P reduced temperature and pressure of the gas can be obtained using the ratios Tr  and Pr  Tc Pc Tr 

TCH4 TcCH

4

500.0 K 190.6 K Tr  2.623

Tr 

Pr  Pr 

PCH4 PcCH

4

2.00 bar 46.0 bar

Pr  4.35  102 TH2 and PH2 are given by rearranging the ratios for reduced temperature and pressure. TH2  TrTcH and PH2  Pr PcH 2

2

Solving for TH2 and PH2 gives;

1-120

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

TH2   2.623 33.2 K  TH2  87.0836 K





PH2  4.35 102 13.0 bar  PH2  0.5655 bar

Tr 

TCH4 TcCH

4

500.0 K 190.6 K Tr  2.623

Tr 

Pr  Pr 

PCH4 PcCH

4

2.00 bar 46.0 bar

Pr  4.35 102 TcH  33.2 K 2

PcH  13.0 bar 2

TH2  87.1 K PH2  0.566 bar

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1-121

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.59. For the Dieterici equation, derive the relationship of a and b to the critical volume and temperature. [Hint: Remember that at the critical point (∂P/∂V)T = 0 and (∂2P/∂V2)T = 0.]

Solution: Given: Dieterici equation: ( Pe a / Vm RT )(Vm  b)  RT Required: Tthe relationship of a and b to Vc and Tc By rearranging for P and using Eq.1.114, (∂P/∂V)T and (∂2P/∂V2)T can be determined. P

RT e  a /Vm RT V b   m 

 a  1    a /Vm RT RT  P  RT  a /Vm RT e         2   e 2 Vm  b  Vm  b   Vm T  RT  Vm    a   P  1 RT e  a /Vm RT       2 Vm  b   Vm T  RTVm Vm  b    a   P  1      P  2  Vm T  RTVm Vm  b  

   2 P   P   a 1   2a 1       P       2 2 2 3    Vm T  Vm T  RTVm Vm  b    RTVm Vm  b   2

   a  2 P  1  2a 1 P P           2  2    RTVm 3 V  b 2   Vm T  RTVm Vm  b   m   2  a   2P  1  2a 1   P        2 2  2 3 V RTV V b RTV       V b    m m m  m T  m  

1-122

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Solving for (∂P/∂V)T = 0 and (∂2P/∂V2)T = 0, the condition of the critical point, The relationship of a and b to Vc and Tc can then be obtained;  a   P  1    0    P  2  Vm Tc  RTVm Vm  b  

 a 1  P     0 2  RTcVc Vc  b   a

RTcVc 2 Vc  b 

2  a   2 P  1  2a 1  0  P       2 2  2 3  RTVm Vm  b   RTVm Vm  b    Vm T   2  a  1  2a 1  0 P      2 2 3  RTcVc Vc  b   RTcVc Vc  b     2

 a 1  2a 1   0    2 2 3  RTcVc Vc  b   RTcVc Vc  b 

Substituting the expression for a into the above can further simplify the problem;

1-123

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

 1   RTcVc 2 

2

 RTcVc 2  1  2     Vc  b   Vc  b   RTc Vc 3   

 1 1    V b V b  c   c  2 1  Vc Vc  b 

Solutions

 RTcVc 2  1   0  Vc  b   Vc  b 2  

2

 2 1    0  Vc Vc  b  V  b 2 c 

2 Vc  b   Vc b  Vc  b

Vc 2

Vc 2

The Dieterici constant a then becomes, a

RTcVc 2 Vc    Vc   2 

RTcVc 2 a  Vc    2 a  2 RTcVc

Substitution back into the Dieterici equation, the critical point becomes, P

V RT e  a /Vm RT , a  2 RTcVc and b  c 2 Vm  b 

1-124

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

2 RTcV

Solutions

c

 RTc V RT Pc  e c c Vc    Vc   2  RTc 2 Pc  e  Vc   2  

Pc 

2 RTc 2 e Vc

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1-125

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.60. In Eq. 1.103 a cubic equation has to be solved in order to find the volume of a van der Waals gas. However, reasonably accurate estimates of volumes can be made by deriving an expression for the compression factor Z in terms of P from the result of the previous problem. One simply substitutes for the terms Vm on the right-hand side in terms of the ideal gas law expression Vm = RT/P. Derive this expression and use it to find the volume of CCl2F2 at 30.0 °C and 5.00 bar pressure. What will be the molar volume computed using the ideal gas law under the same conditions? Solution: Given: (from problem 1.59): CCl2F2 at T  30.0 C = 303.15 K and P = 5.00 bar Required: Vm and Vmideal The compression factor ; Z  P

PV PVm  can be used with Eq. 1.101 to obtain an expression for Z; nRT RT

RT a  2 Vm  b Vm

Z 

 Vm PVm  Vm  RT    RT  RT  Vm  b  RT

Z 

Vm a  Vm  b RTVm

 a  2  Vm 

Since Vm/(Vm – b) = (1 – b/Vm)–1, and (1 – x)–1 = 1 + x + x2 + …, therefore; –1

2

 b  b  b      1 –   1 Vm  Vm  Vm   2

b  b  a 1     Z  1 Vm  Vm  RT Vm 1  a Z  1  b  Vm  RT

2

  b         Vm 

Using Vm = RT/P, we obtain; 1-126

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Z  1

P RT

a  b  RT 

  b    RT

Solutions

 2 P 

Table 1.5 gives the van der Waals constants for CCl2F2 : a  1.066 dm3 bar mol2 , b  0.0973 dm3 mol1 Z  1

 5.00

 0.083 15 bar dm

3

bar





K –1 mol –1 303.15 K





  1.066 dm 3 bar mol2  0.0973 dm3 mol1    3 –1 –1 303.15 K  0.083 15 bar dm K mol  





 0.0973 dm 3 mol1   0.083 15 bar dm 3 K –1 mol –1  Z  1.01167 Z  1.01





2



  5.00 bar 303.15 K  







2

We can then solve for Vm by rearranging the expression for the compression factor;

PVm RT Z RT Vm  P Z

Vm 





1.01 0.083 15 bar dm3 K –1 mol –1 303.15 K

 5.00 bar 



Vm  5.0918dm3 mol –1 Vm  5.09dm3 mol –1

1-127

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Vmideal can also be obtained using the ideal gas, PV = nRT , and solving for Vm,

Vmideal  Vmideal

RT P

 0.083 15 bar dm 

3



K –1 mol –1 303.15 K

 5.00 bar 

  

Vmideal  5.04138 dm3 mol –1 Vmideal  5.04 dm3 mol –1    

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1-128

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.61. A general requirement of all equations of state for gases is that they reduce to the ideal gas equation (Eq. 1.28) in the limit of low pressures. Show that this is true for the van der Waals equation. Solution: Given: PV = nRT, low P Required: show that the van der Waals equation reduces to Eq. 1.28 The Van der Waals equation is given by;  an 2  P  (V  nb)  nRT  V 2 

(Eq. 1.100)

Using the compression factor, Z  P

PV PVm  , Eq. 1.100 can be recast in a form of Z in terms of P. nRT RT

RT a  2 Vm  b Vm

Z 

 Vm PVm  Vm  RT    RT  RT  Vm  b  RT

Z 

Vm a  Vm  b RTVm

 a  2 V  m

Since Vm/(Vm – b) = (1 – b/Vm)–1, and (1 – x)–1 = 1 + x + x2 + …, therefore; –1

2

 b  b  b      1 –   1 V V m  m   Vm  2

b  b  a 1 Z  1     Vm  Vm  RT Vm 1  a Z  1 b  Vm  RT

2

  b         Vm  1-129

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

Using Vm = RT/P we obtain; Z  1

P RT

a  b  RT 

  b    RT

 2 P 

Taking the limit of Z as P approaches 0 becomes,  P  a   b  2 lim 1  b   P  P 0 RT  RT   RT   lim  1

  

P 0

Z 1 Which is true for an ideal gas, and therefore the van der Waals equation reduces to the ideal gas equation.

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1-130

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.62. The van der Waals constants for C2H6 in the older literature are found to be a = 5.49 atm L2 mol–2 and b = 0.0638 L mol–1 Express these constants in SI units (L = liter = dm3).

Solution: Given: PV = nRT, low P, a = 5.49 atm L2 mol–2 and b = 0.0638 L mol–1 Required: express a and b in SI units 1atm  101 325 Pa 1 L2  1 dm    0.1 m   1106 m 6 6

6

a  5.49 atm L2 mol –2 

101 325 Pa 1106 m 6  1atm 1 L2

a  5.56 101 Pa m 6 mol –2 1 L  1 dm    0.1 m   1103 m3 3

3

1 103 m3 b  0.0638 L mol  1L –1

b  6.38  105 m3 mol –1

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1-131

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.63. Compare the values obtained for the pressure of 3.00 mol CO2 at 298.15 K held in a 8.25-dm3 bulb using the ideal gas, van der Waals, Dieterici, and Beattie-Bridgeman equations. For CO2 the Dieterici equation constants are a = 0.462 Pa m6 mol–2, b = 4.63 × 10–5 m3 mol–1

Solution: Given: n  3.00 mol, TCO2  298.15 K, VCO2  8.25 dm3 aDieterici  0.462 Pa m 6 mol –2 , bDieterici  4.63  10 –5 m3 mol –1 Required: Pideal , Pvdw , PDieterici and PBB The Ideal Gas equation is given by; Pideal  Pideal 

Pideal

nRT V

 3.00 mol  8.3145 J K

1

mol1

  298.15 K 

 1 103 m3  3 8.25 dm    1 dm3    901 400 J m -3

where 1 J  1 kg m 2 s 2 , 1 Pa  1 kg m 1 s 2 Pideal  901 400 kg m 2 s 2 m 3 Pideal  901 400 Pa where 1 bar  100 000 Pa Pideal  9.01 bar The Van der Waals equation is defined as;

1-132

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Pvdw 

Solutions

RT a  2 Vm  b Vm

V  1 103 m3  1   8.25 dm3    3  n  1 dm   3.00 mol  Vm  2.75  103 m3 mol1

Vm 

From Table 1.5, a  0.3640 Pa m 6 mol –2 , b  0.0427  10 –3 m3 mol –1 Pvdw 

8.3145 J K

 2.75 10

3

-1

mol-1

 

  298.15 K 

m3 mol1  0.0427  10 –3 m3 mol –1

 0.3640 Pa m mol     2.75 10 m mol  6

3

–2

1

3

2

Pvdw  915 661 J m 3  48132 Pa Pvdw  915 661 Pa  48132 kg m 1 s 2 Pvdw  867 528 Pa Pvdw  8.68 bar

The Dieterici Equation is as follows; PDieterici 

RT e  a /Vm RT Vm  b 

aDieterici  0.462 Pa m 6 mol –2 , bDieterici  4.63  10 –5 m3 mol –1 , Vm  2.75 103 m3 mol1 PDieterici 

8.3145 J K

 2.75 10

3

1

mol



1

  298.15 K 

m3 mol-1  4.63  10 –5 m3 mol –1

 0.462 Pa m mol  mol  8.3145 J K mol  298.15 K  6





e

 2.7510

3

m3

1

–2

1

1

PDieterici  856 801 Pa PDieterici  8.57 bar

The Beattie-Bridgeman equation; 1-133

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

  c  A V  B  2 1   3   m Vm   VmT     a  b  where A  A0 1   , B  B0 1    Vm   Vm  PBB 

RT Vm2

From Table 1.6: A0  0.50728Pa m 6 mol –2 , a  71.32 10 –6 m 3 mol –1 , B0  104.76 10 –6 m 3 mol –1 , b  72.35  10 –6 m3 mol –1 , c  66.00 10 m3 K 3 mol –1  71.32  10 –6 m3 mol –1 A  0.50728 Pa m 6 mol –2 1   2.75 103 m3 mol –1  A  0.494124 Pa m6 mol –2

   

 72.35 10 –6 m3 mol –1 B  104.76  10 –6 m3 mol –1 1   2.75  103 m3 mol –1  B 1.02004  10 –4 m3 mol –1





   

Solving for PBB gives,

PBB



8.3145 J K  

 2.75 10





mol –1 298.15 K   66.00  10 m3 K 3 mol –1  1  2   3 3 –1 2.75 103 m3 mol –1 298.15 K   2.75 10 m mol

3

1

–1





–4

3

m mol

–1



 0.013156 Pa m mol     2.75 10 m mol  6

m mol  1.02004  10 3





  3   

3

3

–2

–1

2

PBB  861 075 Pa PBB  8.61 bar

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1-134

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.64. A gas obeys the van der Waals equation with Pc = 3.040 × 106 Pa (= 30 atm) and Tc = 473 K. Calculate the value of the van der Waals constant b for this gas. Solution: Given: Pc = 3.040 × 106 Pa (=30 atm) and Tc = 473 K Required: b From Eq. 1.109; b 

Vc 8PV and R  c c 3Tc 3

It is possible to rearrange the expression for the gas constant to express it in terms of Vc, so that we can isolate for b.

Vc 

3RTc 8Pc

1 3 RTc 3 8Pc RT b c 8Pc

b

8.3145 J K b

1



mol1 473 K

8  3.040 10 Pa  6

where 1 J  1 kg m2 s2 and 1 Pa  1 kg m1 s 2 1 kg m 2 s 2 1J   1 m3 1 Pa 1 kg m 1 s 2 b 1.617 09 104 m3 mol1 b 1.62 104 m3 mol1

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1-135

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

1.65. Expand the Dieterici equation in powers of Vm1 in order to cast it into the virial form. Find the second and third virial coefficients. Then show that at low densities the Dieterici and van der Waals equations give essentially the same result for P. Solution: Given: ( Pe a / Vm RT )(Vm  b)  RT Required: second and third virial coefficients First, the Dieterici equation can be rewritten in terms of P as; PDieterici 

RT e  a /Vm RT Vm  b 

The series expansion for ex is given by e x  1  x 

PDieterici

x 2 x3   ... can be used to expand the Dieterici equation. 2! 3!

2    a      RT   a   Vm RT      ... 1  2! Vm  b    Vm RT       2

PDieterici PDieterici

 a   RT   a   RT   Vm RT  RT     ...   Vm  b   Vm  b    Vm RT   Vm  b   2! RT a a2     ... Vm  b  Vm Vm  b  2 RTVm 2 Vm  b 

Expanding

1 and collecting terms in powers of Vm gives coefficients that are independent of Vm. : Vm  b

1-136

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases

Solutions

–1 2  1 1  b  1  b  b  1       1 –    Vm  b Vm  Vm  Vm  Vm  Vm   

Substitution into the Dieterici equation leads to; P

 RT a  RTb 1  a 2   3  ab  RTb 2   ... 2 Vm Vm Vm  2 RT 

 a2   ab  RTb 2  . The second coefficient is   a  RTb  and the third coefficient is   2 RT  At low densities, the third and higher terms are negligible. Dropping the third and higher terms, and substituting, we obtain P

RT a  RTb  Vm Vm 2

This is in the same form as the van der Waals equation.

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1-137